BÀi 4:tìm x, biết:
a, 3(2x -1)(x + 2) – 2(3x + 2)(x – 4) = 5
b, (3x + 2)(2x + 9) – (x + 2)(6x + 1) = (x + 1) – (x – 6)
c, (x + 1)(x + 2) – (x – 3)(x + 4) = 6
d, 3(2x – 1)(3x – 1) – (2x – 3)(9x – 1) = 0
e, (2x-1)(3 – x) + (x – 2)(x + 3) =(1 – x)(x – 2)
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BÀi 4:tìm x, biết: a, 3(2x -1)(x + 2) – 2(3x + 2)(x – 4) = 5 b, (3x + 2)(2x + 9) – (x + 2)(6x + 1) = (x + 1) – (x – 6) c, (x + 1)(x + 2) – (x – 3)(x +
By Gabriella
Đáp án:
bai4:
a,3(2x-1)(x+2)-2(3x+2)(x-4)=5
<=> 3( $2x^{2}$+3x-2)-2( $3x^{2}$-12x+2x-8)=5
<=> $6x^{2}$+12x-3x-6-$6x^{2}$ -24x+4x-16=5
<=>-11x=5+22
<=>-11x=27
<=>x=$\frac{-27}{11}$
b,(3x+2)(2x+9)-(x-2)(6x+1)=(x+1)-(x-6)
<=>$6x^{2}$+27x+4x+18- ($6x^{2}$+x-12x-2)=X+1-x+6
<=>$6x^{2}$+31x+18- 6$x^{2}$+11x+2=7
<=>42x+20=7
<=>42x=-13
<=>x=$\frac{-13}{42}$
bai6:
c,(x+1)(x+2)-(x-3)(x+4)=6
<=>$x^{2}$+2x+x+2-($x^{2}$+4x-3x-12)=6
<=>$x^{2}$+3x+2-$x^{2}$-x+12=6
<=>2x+14=6
<=>2x=20
<=>10
d,3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>3(6$x^{2}$-5x+1)-($18x^{2}$-29x+3)=0
<=>$18x^{2}$-15x+3-$18x^{2}$+29x-3=0
<=>14x=3-3
<=>14x=0
<=>x=0
vo nghiem
e,(2x-1)(3 – x) + (x – 2)(x + 3) =(1 – x)(x – 2)
<=>6x-$2x^{2}$-3+x+$x^{2}$+3x-2x-6=x-2-$x^{2}$+2x
<=>-$x^{2}$+8x-9=-$x^{2}$+3x-2
<=>8x-3x=-2+9
<=>5x=7
<=>x= $\frac{7}{5}$
Giải thích các bước giải:kh hieu hoi minh nha
a) 3(2x -1)(x + 2) – 2(3x + 2)(x – 4) = 5
3.(2x²+4x-x-2) – 2.(3x²-12x+2x-8)=5
6x²+12x-3x-6 -6x²+24x-4x+16 = 5
29x = -5
x= -5/29
b) (3x + 2)(2x + 9) – (x + 2)(6x + 1) = (x + 1) – (x – 6)
6x²+27x+4x+18-6x²-x-12x-2 = x+1-x+6
27x+4x-x-12x-x+x = -18+2+1+6
18x = 27
x = 3/2
c) (x + 1)(x + 2) – (x – 3)(x + 4) = 6
x²+2x+x+2 -x²-4x+3x-12 =6
2x+x-4x+3x = -2+12+6
2x = 16
x = 8
d) 3(2x – 1)(3x – 1) – (2x – 3)(9x – 1) = 0
3.(6x²-2x-3x+1)-18x²+2x+27x-3 =0
18x²-6x-9x+3 – 18x²+2x+27x-3 =0
-6x-9x+2x+27x = -3+3
14x =0
x = 0
e) (2x-1)(3 – x) + (x – 2)(x + 3) =(1 – x)(x – 2)
6x-2x²-3+x +x²+3x-2x-6 = x-2-x²+2x
-2x²+x²+x²+6x+x+3x-2x-2x= 3+6-2
6x= 7
x= 7/6