\(\Leftrightarrow20\left(x^2-4x+3\right)-24\left(4x^2-4x+1\right)=15\left(9x^2+6x+1\right)+90x\left(x-1\right)\)

\(\Leftrightarrow20x^2-80x+60-96x^2+96x-24=135x^2+90x+15+90x^2-90x\)

\(\Leftrightarrow-301x^2+16x+21=0\)

\(\text{Δ}=16^2-4\cdot\left(-301\right)\cdot21=25540\)

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

\(\left\{{}\begin{matrix}x_1=\dfrac{-16-\sqrt{25540}}{-602}=\dfrac{16+\sqrt{25540}}{602}\\x_2=\dfrac{16-\sqrt{25540}}{602}\end{matrix}\right.\)

1) \(x=\frac{99}{196}\)

2) \(x=-2\)

3) \(x\approx-0,59\)

giup mk giải rõ dc ko

Bài 1:

a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)

\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)

\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)

Suy ra: \(12x-45-12x^2+45x=0\)

\(\Leftrightarrow-12x^2+57x-45=0\)

\(\Leftrightarrow-12x^2+12x+45x-45=0\)

\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)

\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)

mà \(-3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)

b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)

\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)

Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)

\(\Leftrightarrow x^2-13x-3x+39=0\)

\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)

Vậy: Tập nghiệm S={3;13}

c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)

\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)

\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)

Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)

\(\Leftrightarrow-21x^2+26x+11=0\)

\(\Leftrightarrow-21x^2-7x+33x+11=0\)

\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)

\(\frac{\left(2-3x\right)^2}{3}-\frac{\left(1+2x\right)^2}{2}=\frac{3}{4}-2\left(x-1\right)\left(x+2\right)+x\left(1+x\right)\)

\(\frac{2^2-12x-3x^2}{3}-\frac{1^2+4x+2x^2}{2}=\frac{3}{4}-\left(x^2+x-2\right)+3x\)

\(\frac{2.\left(4-12x-3x^2\right)}{6}-\frac{3.\left(1+4x+2x^2\right)}{6}=\frac{11}{4}-x^2+2x\)

\(\frac{8-24x-6x^2}{6}-\frac{3+12x+2x^2}{6}=\frac{11}{4}-x^2+2x\)

\(\frac{8-24x-6x^2-3-12x-2x^2}{6}=\frac{11}{4}-x^2+2x\)

\(\frac{5-36x-8x^2}{6}=\frac{11}{4}-x^2+2x\)

Chỗ đây thì mk chịu

\(\Leftrightarrow\dfrac{1}{2}\left(x^2-4x+4\right)-\dfrac{13}{3}\left(x^2+6x+9\right)=\dfrac{1}{4}\left(x^2-3x+2\right)-2\left(9x^2+3x-2\right)\)

\(\Leftrightarrow x^2\cdot\dfrac{1}{2}-2x+2-\dfrac{13}{3}x^2-26x-39=\dfrac{1}{4}x^2-\dfrac{3}{4}x+\dfrac{1}{2}-18x^2-6x+4\)

\(\Leftrightarrow x^2\cdot\dfrac{167}{12}-\dfrac{85}{4}x-\dfrac{83}{2}=0\)

\(\Leftrightarrow167x^2-255x-498=0\)

\(\text{Δ}=\left(-255\right)^2-4\cdot167\cdot\left(-498\right)=397689\)

Vì Δ>0 nên phương trình có 2 nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{255-\sqrt{397689}}{334}\\x_2=\dfrac{255+\sqrt{397689}}{334}\end{matrix}\right.\)

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

máy tính tính dc vô nghiệm

Giải ra mk vs

Giải ra mk vs

​

\(\Leftrightarrow\dfrac{1}{3}\left(4x^2-4x+1\right)-\dfrac{1}{2}\left(9x^2+6x+1\right)=\dfrac{1}{3}\left(2x-3x^2-2+3x\right)\)

\(\Leftrightarrow\dfrac{4}{3}x^2-\dfrac{4}{3}x+\dfrac{4}{3}-\dfrac{9}{2}x^2-3x-\dfrac{1}{2}=\dfrac{1}{3}\left(-3x^2+5x-2\right)\)

\(\Leftrightarrow x^2\cdot\dfrac{-19}{6}-\dfrac{13}{3}x+\dfrac{5}{6}+x^2-\dfrac{5}{3}x+\dfrac{2}{3}=0\)

\(\Leftrightarrow x^2\cdot\dfrac{-13}{6}-6x+\dfrac{3}{2}=0\)

\(\text{Δ}=\left(-6\right)^2-4\cdot\left(-\dfrac{13}{6}\right)\cdot\dfrac{3}{2}=49\)

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

\(\left\{{}\begin{matrix}x_1=\dfrac{6-7}{2\cdot\dfrac{-13}{6}}=\dfrac{3}{13}\\x_2=\dfrac{6+7}{2\cdot\dfrac{-13}{6}}=-3\end{matrix}\right.\)