lop 8 con lic le 3

\(\left(2x+1\right)\left(4x^2-2x+1\right)-8x\left(x^2+2\right)=17\\ \Leftrightarrow\left(8x^3-4x^2+2x+4x^2-2x+1\right)-\left(8x^3+16x\right)=17\\ \Leftrightarrow\left(8x^3+1\right)-\left(8x^3+16x\right)=17\\ \Leftrightarrow16x+1=17\\ \Leftrightarrow x=1\)

1)  (2x + 1)(3x – 2) = (5x – 8)(2x + 1)

⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0

⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0

⇔ (2x + 1).(3x – 2 – 5x + 8) = 0

⇔ (2x + 1)(6 – 2x) = 0

⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)

Vậy.....

2)  4x² -1 = (2x + 1)(3x - 5)

⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0

⇔ (2x+1)(2x-1-3x+5)=0

⇔ (2x+1)(4-x)=0

⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)

Vậy...

3)  

(x + 1)² = 4(x² – 2x + 1)

⇔ (x + 1)² - 4(x² – 2x + 1) = 0

⇔ x² + 2x +1- 4x² + 8x – 4 = 0

⇔ - 3x² + 10x – 3 = 0

⇔ (- 3x² + 9x) + (x – 3) = 0

⇔ -3x (x – 3)+ ( x- 3) = 0

⇔ ( x- 3) ( - 3x + 1) = 0

⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy......

4) 2x³+5x²-3x=0

⇒2x³-x²+6x²-3x=0

⇒(2x³-x²)+(6x²-3x)=0

⇒x²(2x-1)+3x(2x-1)=0

⇒(x²+3x)(2x-1)=0

⇒ hoặc x²+3x=0⇒x(x+3)=0⇒hoặc x=0 hoặc x=-3

hoặc 2x-1=0⇒x=0,5

Vậy ...

5)2x=3x-2

⇒2x-3x=-2

⇒-x=-2

⇒x=2

6) x+15=3x-1

⇒x-3x=-1-15

⇒-2x=-16

⇒x=8

7)2-x=0,5x-4

⇒-x-0,5x=-4-2

⇒-1,5x=-6

⇒x=4

tìm x biết:

(3x-1) [- 1/2x+5]=0

1/4+1/3:(2x-1)=-5

[2x+3/5]² - 9/25=0

-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6

[x+1/2]x [2/3-2x]=0

17/2-|2x-3/4|=-7/4

2/3x-1/2x =5/12

(x+1/5)²+17/25=26/25

[x.44/7+3/7].11/5-3/7=-2

3[3x-1/2]+1/9=0

Toán lớp 6Tìm x

 Trả lời  Câu hỏi tương tự

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3(3x-1/2)^3+1/9=0

\(\Leftrightarrow10x-5-2+2x=17\\ \Leftrightarrow8x=24\\ \Leftrightarrow x=3\)

Lời giải:
PT $\Leftrightarrow \frac{11}{17}x+\frac{3}{17}=\frac{78}{17}x-\frac{22}{17}$

$\Leftrightarrow \frac{25}{17}=\frac{67}{17}x$

$x=\frac{25}{67}$

\(x-\dfrac{3}{17}\left(2x-1\right)=\dfrac{7}{34}\left(1-2x\right)+\dfrac{10x-3}{2}\)

<=>\(x-\dfrac{3}{17}\left(2x-1\right)-\dfrac{7}{34}\left(1-2x\right)=\dfrac{10x-3}{2}\)

<=>\(x-\dfrac{3}{17}\left(2x-1\right)+\dfrac{7}{34}\left(2x-1\right)=\dfrac{10x-3}{2}\)

<=>\(x+\left(-\dfrac{3}{17}+\dfrac{7}{34}\right)\left(2x-1\right)-\dfrac{10x-3}{2}=0\)

<=>\(x+\dfrac{1}{34}\left(2x-1\right)-\dfrac{10x-3}{2}=0\)

<=>\(x+\dfrac{1}{17}x-\dfrac{1}{34}-\dfrac{10x-3}{2}=0\)

<=>\(\dfrac{18}{17}x-\dfrac{85x-25.5}{17}=\dfrac{0.5}{17}\)

=> \(18x-85x+25.5=0.5\)
<=>\(-67x=0.5-25.5\)
<=>\(-67x=-25\)
<=> \(x=\dfrac{25}{67}\)
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