Bài toán 2. Phân tích thành nhân tử a) A = (a – b)^2 – c^2 b) B = (m + n)^2 – (m – n)^2 c) C = a^4b^2 – x^8 d) D = (2a + b)^2 – (2b + a)^2 e) E = (a

Question

Bài toán 2.
Phân tích thành nhân tử
a) A = (a – b)^2 – c^2
b) B = (m + n)^2 – (m – n)^2
c) C = a^4b^2 – x^8
d) D = (2a + b)^2 – (2b + a)^2 e) E = (a + b)^2 – (b + c)^2
Bài toán 3.
Phân tích thành nhân tử
a) A = x^3 – y^3
b) B = x^3 + 27
c) c = a^3 – 27
d) D = 125 – b^3
e) E = m^3 – 64
Bài toán 4.
Phân tích thành nhân tử
a) A = 27a^3 – 8
b) B = (8a^3 – 27b^3) – 2a(4a^2 – 9b^2)
c) C = (a^3 – b^3) + (a – b)^2
d) D = (a^3 + b^3) + (a + b)^2
Bài toán 5.
Phân tích thành nhân tử
a) A = (a^2 + 1)^2 – 4a^2
b) B = (x^2 + 4)^2 – 16x^2
c) C = (a^2 + 2ab + b^2) – c^2
d) D = 1 – (x^2 – 2xy + y^2)
e) E = 2x^2 + 2y^2 – 4xy.
II. TÌM GIÁ TRỊ CỦA BIẾN
Bài toán 6.
Tìm x, biết:
a) x^2 – 36 = 0
c) 25 – l0x + x^2 = 0
b) 4x^2 + 4x +1 = 0
d) x^3 + 8 = 0
Bài toán 7.
Tìm x, biết:
a) x^3 – 3x^2 + 3x – 1 = 0 (1)
c) x^6 – 1 = 0
b) 4x^3 – 36x = 0 (2)
d) x^3 – 6x^2 + 12x – 8 = 0
Bài toán 8.
Tìm x, biết:
a) 9x^2 – 16(x – 1)^2 = 0
b) (5x – 4)^2 – 49x^2 = 0
e) (x^3 + 9) – (x + 2)(x – 4)
b) 4x^3 – 36x = 0 (2)
d) x^3 – 6x^2 + 12x – 8 = 0 (4)

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Claire 2 tuần 2021-07-12T12:28:27+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-12T12:29:46+00:00

    Đáp án:

     

    Giải thích các bước giải:

     

    0
    2021-07-12T12:29:50+00:00

    Đáp án+Giải thích các bước giải:

     Bài 2:

    `a)`

    `A=(a-b)^2-c^2`

    `=(a-b-c).(a-b+c)`

    `b)`

    `B=(m+n)^2-(m-n)^2`

    `=(m+n+m-n).(m+n-m+n)`

    `=2m.2n`

    `=4.m.n`

    `c)`

    `C=a^4.b^2-x^8`

    `=(a^2.b)^2-(x^4)^2`

    `=(a^2.b-x^4).(a^2.b+x^4)`

    `d)`

    `D=(2a+b)^2-(2b+a)^2`

    `=(2a+b-2b-a).(2a+b+2b+a)`

    `=(a-b).(3a+3b)`

    `=3.(a+b).(a-b)`

    `e)`

    `E=(a+b)^2-(b+c)^2`

    `=(a+b+b+c).(a+b-b-c)`

    `=(a+2b+c).(a-c)`

    Bài 3:

    `a)`

    `A=x^3-y^3`

    `=(x-y).(x^2+xy+y^2)`

    `b)`

    `B=x^3+27`

    `=x^3+3^3`

    `=(x+3).(x^2-3x+9)`

    `c)`

    `C=a^3-27`

    `=a^3-3^3`

    `=(a-3).(a^2+3a+9)`

    `d)`

    `D=125-b^3`

    `=5^3-b^3`

    `=(5-b).(25+5b+b^2)`

    `e)`

    `E=m^3-64`

    `=m^3-4^3`

    `=(m-4).(m^2+4m+16)`

    Bài 4:

    `a)`

    `A=27a^3-8`

    `=(3a)^3-2^3`

    `=(3a-2).(9a^2+6a+4)`

    `b)`

    `B=(8a^3-27b^3)-2a(4a^2-9b^2)`

    `=[(2a)^3-(3b)^3]-2a.[(2a)^2-(3b)^2]`

    `=(2a-3b).(4a^2+6ab+9b^2)-2a.(2a+3b).(2a-3b)`

    `=(2a-3b).(4a^2+6ab+9b^2-2a.(2a+3b))`

    `=(2a-3b).(4a^2+6ab+9b^2-4a^2-6ab)`

    `=(2a-3b).9.b^2`

    `c)`

    `C=(a^3-b^3)+(a-b)^2`

    `=(a-b).(a^2+ab+b^2)+(a-b).(a-b)`

    `=(a-b).(a^2+ab+b^2+a-b)`

    `d)`

    `D=(a^3+b^3)+(a+b)^2`

    `=(a+b).(a^2-ab+b^2)+(a+b).(a+b)`

    `=(a+b).(a^2-ab+b^2+a+b)`

    Bài 5:

    `a)`

    `A=(a^2+1)^2-4a^2`

    `=(a^2+1)^2-(2a)^2`

    `=(a^2+1+2a).(a^2+1-2a)`

    `=(a+1)^2.(a-1)^2`

    `b)`

    `B=(x^2+4)^2-16x^2`

    `=(x^2+4)^2-(4x)^2`

    `=(x^2-4x+4).(x^2+4x+4)`

    `=(x-2)^2.(x+2)^2`

    `c)`

    `C=(a^2+2ab+b^2)-c^2`

    `=(a+b)^2-c^2`

    `=(a+b-c).(a+b+c)`

    `d)`

    `D=1-(x^2-2xy+y^2)`

    `=1-(x-y)^2`

    `=(1-x+y).(1+x-y)`

    `e)`

    `E=2x^2+2y^2-4xy`

    `=2.(x^2-2xy+y^2)`

    `=2.(x-y)^2`

    Bài 6:

    `a)`

    `x^2-36=0`

    `<=>x^2-6^2=0`

    `<=>(x-6).(x+6)=0`

    `<=>`\(\left[ \begin{array}{l}x-6=0\\x+6=0\end{array} \right.\)

    `<=>`\(\left[ \begin{array}{l}x=6\\x=-6\end{array} \right.\) 

    Vậy `S={-6;6}`

    `b)`

    `4x^2+4x+1=0`

    `<=>(2x)^2+2.2.x+1=0`

    `<=>(2x+1)^2=0`

    `<=>2x+1=0`

    `<=>x=-1/2`

    Vậy `S={-1/2}`

    `c)`

    `25-10x+x^2=0`

    `5^2-2.5.x+x^2=0`

    `<=>(5-x)^2=0`

    `<=>5-x=0`

    `<=>x=5`

    Vậy `S={5}`

    `d)`

    `x^3+8=0`

    `<=>x^3+2^3=0`

    `<=>(x+2).(x^2-2x+4)=0`

    `<=>`\(\left[ \begin{array}{l}x+2=0\\x^2-2x+4=0\end{array} \right.\) 

    `+)x^2-2x+4=(x-1)^2+3`

    Vì `(x-1)^2≥0⇒(x-1)^2+3≥3>0`

    `=>` Không có `x` thỏa mãn: `x^2-2x+4=0`

    `+)x+2=0`

    `<=>x=-2`

    Vậy `S={-2}`

    Bài 7:

    `a)`

    `x^3-3x^2+3x-1=0`

    `<=>(x-1)^3=0`

    `<=>x-1=0`

    `<=>x=1`

    Vậy `S={1}`

    `b)`

    `4x^3-36x=0`

    `<=>4.x.(x^2-9)=0`

    `<=>4.x.(x+3).(x-3)=0`

    `<=>`\(\left[ \begin{array}{l}x=0\\x+3=0\\x-3=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=0\\x=-3\\x=3\end{array} \right.\) 

    Vậy `S={0;-3;3}`

    `c)`

    `x^6-1=0`

    `<=>(x^2)^2-1^3=0`

    `<=>(x^2-1).(x^4+x^2+1)=0`

    Vì `x^4+x^2+1\ne0`

    `<=>x^2-1=0`

    `<=>(x-1).(x+1)=0`

    `<=>`\(\left[ \begin{array}{l}x-1=0\\x+1=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\) 

    Vậy `S={-1;1}`

    `d)`

    `x^3-6x^2+12x-8=0`

    `<=>x^3-3.x.x^2+3.2^2.x-2^3=0`

    `<=>(x-2)^3=0`

    `<=>x-2=0`

    `<=>x=2`

    Vậy `S={2}`

    Bài 8:

    `a)`

    `9x^2-16.(x-1)^2=0`

    `<=>(3x)^2-[4.(x-1)]^2=0`

    `<=>[3x-4.(x-1)][3x+4.(x-1)]=0`

    `<=>(3x-4x+4).(3x+4x-4)=0`

    `<=>(4-x).(7x-4)=0`

    `<=>`\(\left[ \begin{array}{l}4-x=0\\7x-4=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=4\\x=\dfrac{4}{7}\end{array} \right.\) 

    Vậy `S={4; 4/7}`

    `b)`

    `(5x-4)^2-49x^2=0`

    `<=>(5x-4)^2-(7x)^2=0`

    `<=>(5x-4-7x).(5x-4+7x)=0`

    `<=>(-2x-4).(12x-4)=0`

    `<=>`\(\left[ \begin{array}{l}-2x-4=0\\12x-4=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{3}\end{array} \right.\) 

    Vậy `S={-2;1/3}`

    `e)`

    `(x^3+9)-(x+2).(x-4)`

    Sai đề ; sửa:

    `(x^3+8)-(x+2).(x-4)=0`

    `<=>(x^3+2^3)-(x+2).(x-4)=0`

    `<=>(x+2).(x^2-2x+4)-(x+2).(x-4)=0`

    `<=>(x+2).(x^2-2x+4-x+4)=0`

    `<=>(x+2).(x^2-3x+8)=0`

    Vì `x^2-3x+8=x^2-2.(3)/(2).x+(9)/(4)+(23)/(4)=(x-3/2)^2+(23)/(4)≥23/4>0∀x`

    `<=>x+2=0`

    `<=>x=-2`

    Vậy `S={-2}`

    `b)` (trùng với `b)` bài `7`)

    `4x^3-36x=0`

    `<=>4.x.(x^2-9)=0`

    `<=>4.x.(x+3).(x-3)=0`

    `<=>`\(\left[ \begin{array}{l}x=0\\x+3=0\\x-3=0\end{array} \right.\) 

    `<=>`\(\left[ \begin{array}{l}x=0\\x=-3\\x=3\end{array} \right.\) 

    Vậy `S={0;-3;3}`

    `d)` (trùng với `d)` bài `7`)

    `x^3-6x^2+12x-8=0`

    `<=>x^3-3.x.x^2+3.2^2.x-2^3=0`

    `<=>(x-2)^3=0`

    `<=>x-2=0`

    `<=>x=2`

    Vậy `S={2}`

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