a, <=> x² -2x +1 + 5x -x² =8

<=> 3x +1 =8 

<=> 3x = 7

<=> x= 7/3

b, thiếu đề

c, <=> 2x³ -1 + 2x(4 -x²) = 7

<=> 2x³ + 8x -2³ = 8

<=> 8x =8

<=> x=1

a) \(x^2-4x-7=0\)

Ta có: \(\Delta=4^2+4.28=128,\sqrt{\Delta}=\sqrt{128}\)

pt có 2 nghiệm:

\(x_1=\frac{4+\sqrt{128}}{2}\);\(x_2=\frac{4-\sqrt{128}}{2}\)

b) \(x^2-x-11=0\)

Ta có: \(\Delta=1^2+4.11=45,\sqrt{\Delta}=\sqrt{45}\)

pt có 2 nghiệm:

\(x_1=\frac{1+\sqrt{45}}{2}\)\(x_2=\frac{1-\sqrt{45}}{2}\)

Giúp

\(\dfrac{3}{7}\times x=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}:\dfrac{3}{7}\)

\(x=\dfrac{7}{9}\)

\(x:\dfrac{6}{8}=\dfrac{2}{5}\)

\(x=\dfrac{2}{5}\times\dfrac{6}{8}\)

\(x=\dfrac{6}{20}=\dfrac{3}{10}\)

\(x-\dfrac{2}{3}=\dfrac{5}{6}\)

\(x=\dfrac{5}{6}+\dfrac{2}{3}\)

\(x=\dfrac{9}{6}=\dfrac{3}{2}\)

\(x-\dfrac{2}{5}-\dfrac{4}{35}=\dfrac{1}{7}\)

\(x=\dfrac{1}{7}+\dfrac{4}{35}+\dfrac{2}{5}\)

\(x=\dfrac{23}{35}\)

.

\(\dfrac{3}{7}\times x=\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{1}{3}:\dfrac{3}{7}\)

\(\Leftrightarrow x=\dfrac{7}{9}\)

\(x:\dfrac{6}{8}=\dfrac{2}{5}\)

\(\Leftrightarrow x=\dfrac{2}{5}\times\dfrac{6}{8}\)

\(\Leftrightarrow x=\dfrac{3}{10}\)

\(x-\dfrac{2}{3}=\dfrac{5}{6}\)

\(\Leftrightarrow x=\dfrac{5}{6}+\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

\(x-\dfrac{2}{5}-\dfrac{4}{35}=\dfrac{1}{7}\)

\(\Leftrightarrow x=\dfrac{1}{7}+\dfrac{2}{5}+\dfrac{4}{35}\)

\(\Leftrightarrow x=\dfrac{23}{35}\)

c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

a) \(-\dfrac{2}{5}+\dfrac{5}{6}x=-\dfrac{4}{15}\\ \Leftrightarrow\dfrac{5}{6}x=\dfrac{2}{15}\\ \Leftrightarrow x=\dfrac{4}{25}\)

b) \(\dfrac{2}{3}+\dfrac{7}{4}\div x=\dfrac{5}{6}\\ \Leftrightarrow\dfrac{7}{4}\div x=\dfrac{1}{6}\\ \Leftrightarrow x=\dfrac{7}{24}\)

a: Ta có: \(-\dfrac{2}{5}+\dfrac{5}{6}x=\dfrac{-4}{15}\)

\(\Leftrightarrow x\cdot\dfrac{5}{6}=\dfrac{2}{15}\)

hay \(x=\dfrac{4}{25}\)

b: Ta có: \(\dfrac{7}{4}:x+\dfrac{2}{3}=\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{7}{4}:x=\dfrac{1}{6}\)

hay \(x=\dfrac{21}{2}\)

số điện thoại của bạn là gì ?

a. x² - 6x = -9

<=> x² - 6x + 9 = 0

<=> (x - 3)² = 0

<=> x - 3 = 0

<=> x = 3

b. 2(x + 3) - x² + 3x = 0

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

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

<=> \(\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\) 

a: 9-25=7-x-(25+7)

=>7-x-25-7=-16

=>-x-25=-16

=>x+25=16

hay x=-9

b: \(10+2\left|x\right|=2\cdot\left(3^2-1\right)\)

\(\Leftrightarrow2\left|x\right|=2\cdot8-10=6\)

=>x=3 hoặc x=-3

c: -6x=18

nên x=18:(-6)=-3