\(11\left(x-6\right)=4x+11\)

\(\Leftrightarrow11x-66=4x+11\)

\(\Leftrightarrow-66-11=4x-11x\)

\(\Leftrightarrow-77=-7x\)

\(\Leftrightarrow x=11\)

Vậy x=11

\(11\cdot\left(x-6\right)=4x+11\)

\(11x-66=4x+11\)

\(11x-4x=11+66\)

\(7x=77\)

\(\Rightarrow x=11\)

\(11.\left(x-6\right)=4x+11\)
=) \(11x-66=4x+11\)
=) \(11x-4x=11+66\)
=) \(7x=77\)
=) \(x=77:7=11\)

11.(x-6)=4x+11

11.x-66=4x+11

7x=66+11

7x=77

x=11 

vậy x=11

c) (4x - 8)[x + (-3)] = 0

=> 4(x - 2)(x - 3) = 0

=> (x - 2)(x - 3) = 0

=> x - 2 = 0 hoặc x - 3 = 0

+) x - 2 = 0 => x = 2

+) x - 3 = 0 => x = 3

Vậy x \(\in\){2;3}

11(x - 6) = 4x + 11

=> 11x - 66 = 4x + 11

=> 11x - 4x = 11 + 66

=> 7x = 77

=> x = 77/7

=> x = 11

a)11(x-6)=4x+11

11x - 11*6=4x+11

11x-4x=11+11*6

7x=77

x=77/7

x=11

b)lx-3l+1=x

lx-3l=x-1

\(\Rightarrow x-3\in\left\{-\left(x-1\right);x-1\right\}\)

Ta có:

TH1:x-3=-(x-1)

x-3=-x+1

x-(-x)=1+3

2x=4

x=4/2

x=2

TH2:x-3=x-1

x-x=3-1

0=2

\(x\in rỗng\)

Vậy x=2

\(-\dfrac{3}{4}.\dfrac{4}{11}+\left(-\dfrac{5}{4}\right).\dfrac{4}{11}=\dfrac{4}{11}\left(-\dfrac{3}{4}-\dfrac{4}{4}\right)=\dfrac{4}{11}.\left(-\dfrac{4}{7}\right)=-\dfrac{16}{77}\)

\(\dfrac{7}{12}-\left(-\dfrac{1}{5}\right)-\dfrac{5}{6}+\dfrac{2}{3}+\left(-\dfrac{1}{5}\right)\)

\(=\dfrac{7}{12}+\dfrac{1}{5}-\dfrac{5}{6}+\dfrac{2}{3}-\dfrac{1}{5}\)

\(=\dfrac{7}{12}-\dfrac{5}{6}+\dfrac{2}{3}=\dfrac{7}{12}-\dfrac{10}{12}+\dfrac{8}{12}=\dfrac{5}{12}\)

giúp với

Lhoong biet

`2//(5x-8)-3(4x-5)=4(3x-4)`

`<=>5x-8-12x+15=12x-16`

`<=>-19x=-23`

`<=>x=23/19`     Vậy `x=23/19`

`3//2(x^3-1)-2x^2(x+2x^4)+(4x^5+4)x=6`

`<=>2x^3-2-2x^3-4x^6+4x^6+4x=6`

`<=>4x=8`

`<=>x=2`     Vậy `x=2`

`A`

a)

<=> 66 - 11x = 4x + 11

<=> -11x - 4x = 11 - 66

<=> -15x        = -55

<=>    x          = -55 : (-15)

<=>   x           = 11/3

b)

x = 15 . 8 : 6 = 20

5/ (x² – 4) + (x – 2)(4 – 2x) = 0

⇔(x-2)(x+2)+(x – 2)(4 – 2x)=0

⇔(x-2)(x+2+4-2x)=0

⇔(x-2)(6-x)=0

⇔\(\left[{}\begin{matrix}x-2=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

6/ x(2x – 7) – 4x + 14 = 0

⇔2x²-11x+14=0

⇔(x-\(\frac{7}{2}\))(x-2)=0

⇔\(\left[{}\begin{matrix}x-\frac{7}{2}=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)

7/ x² – x – (3x–3)= 0

⇔x²-4x+3=0

⇔(x-3)(x-1)=0

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

8/ (x² – 2x + 1) – 4 = 0

⇔(x-1)²-4=0

⇔(x-1-4)(x-1+4)=0

⇔(x-5)(x+3)=0

⇔\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

9/ 4x² + 4x + 1 = x²

⇔3x²+4x+1=0

⇔(3x+1)(x+1)=0

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

10/ x² – x = - 2x + 2

⇔3x²-x-2=0 (chuyển vế)

⇔(3x+2)(x-1)=0

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

11/ x² – 5x + 6 = 0

⇔x²-3x-2x+6=0

⇔x(x-3)-2(x-3)=0

⇔(x-3)(x-2)=0

⇔\(\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

Mình làm bài khá tắt nên có gì không hiểu bạn cứ hỏi mình nha!

5/(x²-4)+(x-2)(4-2x)=0⇔x²-4+4x-2x²-8+4x=0

⇔-x²+8x-12=0⇔-x²+2x+6x-12=0⇔-x(x-2)-6(x-2)=0⇔(-x-6)(x-2)=0

⇔-x-6=0 hoặc x-2=0⇔x=-6 hoặc x=2