Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

\(\dfrac{1}{6}+x=\dfrac{5}{12}\)
\(=>x=\dfrac{5}{12}-\dfrac{2}{12}=\dfrac{1}{4}\)
\(\dfrac{3}{4}+\dfrac{1}{4}x=-\dfrac{1}{2}\)
\(=>\dfrac{1}{4}x=-\dfrac{5}{4}\)
\(=>x=-\dfrac{5}{4}.4=-5\)
\(7^{2x}+7^{2x+3}=344\)
\(< =>49^x+49^x.343=344\)
\(=>x=?\)

\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)

\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)

\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)

\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)

\(\Rightarrow7x=25x\)

\(\Rightarrow x=0\)

\(\left(4x-3\right)^4=\left(4x-3\right)^2\)

\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)

\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)

Bài mình làm sai à hay sao mà t]ick sai thế???

a: \(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)

=>\(-4x^2+20x-16x+4x^2=-3\)

=>4x=-3

=>\(x=-\dfrac{3}{4}\)

b: \(-7\left(x+9\right)-3\left(5-x\right)=2\)

=>\(-7x-63-15+3x=2\)

=>\(-4x-78=2\)

=>\(-4x=78+2=80\)

=>\(x=\dfrac{80}{-4}=-20\)

1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)

\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)

\(\Leftrightarrow x=2\)

3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)

\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)

\(\Leftrightarrow6x=6\)

hay x=1

a/

\(\left(1-3x\right)^3=\left(-4\right)^3\Leftrightarrow1-3x=-4\Leftrightarrow x=\frac{5}{3}\)

b/

\(\left(4-3x\right)^4=\left(4-3x\right)^2\Leftrightarrow\left(4-3x\right)^2\left[\left(4-3x\right)^2-1\right]=0\)

\(\Leftrightarrow\left(4-3x\right)^2\left(5-3x\right)\left(3-3x\right)=0\)

\(\Leftrightarrow3-3x=0\) hoặc \(4-3x=0\) hoặc \(5-3x=0\)

\(\Leftrightarrow x=1\) hoặc \(x=\frac{4}{3}\) hoặc \(x=\frac{5}{3}\)

c/

\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)\(\Leftrightarrow\frac{49.7^x+7.7^x+7^x}{57}=\frac{5^{2x}+5.5^{2x}+125.5^{2x}}{131}\)

\(\Leftrightarrow7^x=5^{2x}\Leftrightarrow7^x=25^x\Leftrightarrow\left(\frac{7}{25}\right)^x=1=\left(\frac{7}{25}\right)^0\)

\(\Rightarrow x=0\)

\(\left(1-3x\right)^3=-64\)

=> \(1-3x=-4\)

=> \(-3x=-4+1\) (chuyển vế)

=> \(-3x=-3\Rightarrow x=-3:\left(-3\right)=1\)

Tìm x biết
a) (7^x-2+7^x-1+7)/57=(5^x+5^2x+1+5^2x+3)/131
b) (4x-3)^4=(4x-3)^2