Tìm x: 1, 2x^2 = 3x 2, (x-5)^3 = x-5 3, |x-1|+x = 1 4, |x-5|+x=1 08/07/2021 Bởi Skylar Tìm x: 1, 2x^2 = 3x 2, (x-5)^3 = x-5 3, |x-1|+x = 1 4, |x-5|+x=1
`a)` `2x^2=3x` `<=>2x^2-3x=0` `<=>x(2x-3)=0` `<=>` \(\left[ \begin{array}{l}x=0\\2x-3=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=0\\2x=3\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{3}{2}\end{array} \right.\) Vậy `x=0;x=3/2` `b)` `(x-5)^3=x-5` `<=>(x-5)^3-x+5=0` `<=>(x-5)^3-(x-5)=0` `<=>(x-5).[(x-5)^2-1]=0` `<=>(x-5).(x^2-10x+25-1)=0` `<=>(x-5).(x^2-10x+24)=0` `<=>(x-5).(x^2-6x-4x+24)=0` `<=>(x-5).[(x^2-6x)-(4x-24)]=0` `<=>(x-5).[x(x-6)-4(x-6)]=0` `<=>(x-5)(x-6)(x-4)=0` `<=>` \(\left[ \begin{array}{l}x-5=0\\x-6=0\\x-4=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=5\\x=6\\x=4\end{array} \right.\) Vậy `x=5;x=6;x=4` `c)` `|x-1|+x=1` `<=>|x-1|=1-x` ĐKXĐ: `x\leq1` `<=>` \(\left[ \begin{array}{l}x-1=1-x\\x-1=-(1-x)\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x+x=1+1\\x-1=-1+x\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=2\\x-x=-1+1\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=1(\text{thoả mãn})\\0x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1\\x∈R\end{array} \right.\) Vậy `x\leq1` `d)` `|x-5|+x=1` `<=>|x-5|=1-x` Điều kiện: `x\leq1` `<=>` \(\left[ \begin{array}{l}x-5=1-x\\x-5=-(1-x)\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x+x=1+5\\x-5=-1+x\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=6\\x-x=-1+5\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=3(\text{loại})\\0x=4(\text{vô lý})\end{array} \right.\) Vậy ta không tìm được giá trị nào của `x` Bình luận
Đáp án: 1, `2x^2 = 3x` `2x^2 – 3x = 0` `x(2x – 3)=0` ⇒\(\left[ \begin{array}{l}x=0\\2x – 3 =0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=0\\x=3/2\end{array} \right.\) Vậy `x ∈{ 0 ; 3/2}` 2, `(x-5)^3 = x – 5` `(x-5)^3 – (x-5) = 0` `(x-5).((x-5)^2 – 1)=0` ⇒\(\left[ \begin{array}{l}x – 5 = 0\\(x-5)² – 1 = 0\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=5\\x= 4 hoặc x = 6\end{array} \right.\) Vậy` x ∈{ 4 ; 5 ; 6}` 3, `| x – 1| + x = 1` `|x – 1| = 1 – x` ⇒\(\left[ \begin{array}{l}x – 1 = 1 – x\\x – 1 = -1 + x\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=1\\0x = -2( Vô lí)\end{array} \right.\) Vậy `x = 1` 4, `|x-5|+x=1` `| x – 5| = 1 – x` ⇒\(\left[ \begin{array}{l}x – 5 = 1 – x\\x – 5 =-1 + x\end{array} \right.\) ⇒\(\left[ \begin{array}{l}x=3(Loại)\\0x = 4(Vô lí)\end{array} \right.\) Vậy `x ∈ {∅}` Bình luận
`a)` `2x^2=3x`
`<=>2x^2-3x=0`
`<=>x(2x-3)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\2x-3=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=0\\2x=3\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=0\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy `x=0;x=3/2`
`b)` `(x-5)^3=x-5`
`<=>(x-5)^3-x+5=0`
`<=>(x-5)^3-(x-5)=0`
`<=>(x-5).[(x-5)^2-1]=0`
`<=>(x-5).(x^2-10x+25-1)=0`
`<=>(x-5).(x^2-10x+24)=0`
`<=>(x-5).(x^2-6x-4x+24)=0`
`<=>(x-5).[(x^2-6x)-(4x-24)]=0`
`<=>(x-5).[x(x-6)-4(x-6)]=0`
`<=>(x-5)(x-6)(x-4)=0`
`<=>` \(\left[ \begin{array}{l}x-5=0\\x-6=0\\x-4=0\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=5\\x=6\\x=4\end{array} \right.\)
Vậy `x=5;x=6;x=4`
`c)` `|x-1|+x=1`
`<=>|x-1|=1-x` ĐKXĐ: `x\leq1`
`<=>` \(\left[ \begin{array}{l}x-1=1-x\\x-1=-(1-x)\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x+x=1+1\\x-1=-1+x\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=2\\x-x=-1+1\end{array} \right.\) `<=>`\(\left[ \begin{array}{l}x=1(\text{thoả mãn})\\0x=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x=1\\x∈R\end{array} \right.\)
Vậy `x\leq1`
`d)` `|x-5|+x=1`
`<=>|x-5|=1-x` Điều kiện: `x\leq1`
`<=>` \(\left[ \begin{array}{l}x-5=1-x\\x-5=-(1-x)\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x+x=1+5\\x-5=-1+x\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}2x=6\\x-x=-1+5\end{array} \right.\)`<=>` \(\left[ \begin{array}{l}x=3(\text{loại})\\0x=4(\text{vô lý})\end{array} \right.\)
Vậy ta không tìm được giá trị nào của `x`
Đáp án:
1, `2x^2 = 3x`
`2x^2 – 3x = 0`
`x(2x – 3)=0`
⇒\(\left[ \begin{array}{l}x=0\\2x – 3 =0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=0\\x=3/2\end{array} \right.\)
Vậy `x ∈{ 0 ; 3/2}`
2, `(x-5)^3 = x – 5`
`(x-5)^3 – (x-5) = 0`
`(x-5).((x-5)^2 – 1)=0`
⇒\(\left[ \begin{array}{l}x – 5 = 0\\(x-5)² – 1 = 0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=5\\x= 4 hoặc x = 6\end{array} \right.\)
Vậy` x ∈{ 4 ; 5 ; 6}`
3, `| x – 1| + x = 1`
`|x – 1| = 1 – x`
⇒\(\left[ \begin{array}{l}x – 1 = 1 – x\\x – 1 = -1 + x\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=1\\0x = -2( Vô lí)\end{array} \right.\)
Vậy `x = 1`
4, `|x-5|+x=1`
`| x – 5| = 1 – x`
⇒\(\left[ \begin{array}{l}x – 5 = 1 – x\\x – 5 =-1 + x\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=3(Loại)\\0x = 4(Vô lí)\end{array} \right.\)
Vậy `x ∈ {∅}`