a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)

\(3x=5\)

\(x=\frac{5}{3}\)

b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)

\(3x-8=2x-7\)

\(x=1\)

c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)

\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)

\(4x^2-3x-18=4x^2+3x\)

\(6x=-18\)

\(x=-3\)

d) Sai đề

e) ko bt

Mình sẽ làm cách cơ bản cho bạn nhé :)

Ta có : \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)

Đặt \(y=x^2+4x+8\), phương trình trở về dạng \(y^2+3xy+2x^2=0\Leftrightarrow\left(y^2+2xy+y^2\right)+\left(x^2+xy\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+x\left(x+y\right)=0\Leftrightarrow\left(x+y\right)\left(2x+y\right)=0\Leftrightarrow\orbr{\begin{cases}x+y=0\\2x+y=0\end{cases}}\)

Vậy kết luận : Tập nghiệm của phương trình : \(S=\left\{-4;-2\right\}\)

\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)

\(\Rightarrow\left(x^2+4x+8\right)\left(3x+x^2+4x+8\right)+2x^2=0\)

\(\Rightarrow\left(x^2+4x+8\right)\left(x^2+7x+8\right)+2x^2=0\)

Ta có đồng thời :

 \(2x^2=0\Rightarrow x=0\)

Và : \(\left(x^2+4x+8\right)\left(x^2+7x+8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2+4x+8=0\left(vn\right)\\x^2+7x+8=0\end{cases}}\Rightarrow x^2+x-8x+8=0\)

\(\Rightarrow x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Rightarrow\left(x-8\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-8=0\\x-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=1\end{cases}}\)

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

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

\(\Leftrightarrow\left(x^2-1\right)\left(x^2+x-1\right)-2x\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\left(x^2-2x-1\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-2x-1=0\\x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=2\\\left(x+\frac{1}{2}\right)^2=\frac{5}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\pm\sqrt{2}\\x+\frac{1}{2}=\pm\frac{\sqrt{5}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\pm\sqrt{2}\\x=-\frac{1\pm\sqrt{5}}{2}\end{cases}}\)

2) Ta có: \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)

\(\Leftrightarrow\left[\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)\right]+\left[2x\left(x^2+4x+8\right)+2x^2\right]=0\)

\(\Leftrightarrow\left(x^2+4x+8\right)\left(x^2+5x+8\right)+2x\left(x^2+5x+8\right)=0\)

\(\Leftrightarrow\left(x^2+6x+8\right)\left(x^2+5x+8\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)=0\)

Vì \(x^2+5x+8=\left(x^2+5x+\frac{25}{4}\right)+\frac{7}{4}=\left(x+\frac{5}{2}\right)^2+\frac{7}{4}>0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)

Vậy x = -2 hoặc x = -4

P/s: Bạn Thùy Linh nếu PT chứa nghiệm vô tỉ thì với trình độ bình thường không dễ tìm được nghiệm đâu nhé

3) Ta có: \(4\left(x^2+x+1\right)^2+5x\left(x^2+x+1\right)+x^2=0\)

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

\(\Leftrightarrow4\left(x^2+x+1\right)\left(x^2+2x+1\right)+x\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(4x^2+4x+4+x\right)\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(4x^2+5x+4\right)\left(x+1\right)^2=0\)

Xét PT \(4x^2+5x+4=0\) ta có:

\(\Delta_x=5^2-4\cdot4\cdot4=-39< 0\)

\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x=-1\)

Vậy x = -1

\(=64\)

Đặt \(x^2+4x+8=t\)

Khi đó PT có dạng:

\(t^2+3xt+2x^2=t^2-tx-2xt^2+2x^2=t\left(t-x\right)-2x\left(t-x\right)=\left(t-x\right)\left(t-2x\right)\)

\(=\left(x^2+4x+8-x\right)\left(x^2+4x+8-2x\right)=\left(x^2+3x+8\right)\left(x^2+2x+8\right)\)

a) (x-1)(5x+3)=(3x-8)(x-1)

= (x-1)(5x+3)-(3x-8)(x-1)=0

=(x-1)[(5x+3)-(3x-8)]=0

=(x-1)(5x+3-3x+8)=0

=(x-1)(2x+11)=0

\(\Leftrightarrow\) x-1=0 hoặc 2x+11=0

\(\Leftrightarrow\) x=1 hoặc x=\(\dfrac{-11}{2}\)

Vậy S={1;\(\dfrac{-11}{2}\)}

b) 3x(25x+15)-35(5x+3)=0

=3x.5(5x+3)-35(5x+3)=0

=15x(5x+3)-35(5x+3)=0

=(5x+3)(15x-35)=0

\(\Leftrightarrow\) 5x+3=0 hoặc 15x-35=0

\(\Leftrightarrow\) x=\(\dfrac{-3}{5}\) hoặc x=\(\dfrac{7}{3}\)

Vậy S={\(\dfrac{-3}{5};\dfrac{7}{3}\)}

c) (2-3x)(x+11)=(3x-2)(2-5x)

=(2-3x)(x+11)-(3x-2)(2-5x)=0

=(3x-2)[(x+11)-(2-5x)]=0

=(3x-2)(x+11-2+5x)=0

=(3x-2)(6x+9)=0

\(\Leftrightarrow\) 3x-2=0 hoặc 6x+9=0

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

Vậy S={\(\dfrac{2}{3};\dfrac{-3}{2}\)}

d) (2x²+1)(4x-3)=(2x²+1)(x-12)

=(2x²+1)(4x-3)-(2x²+1)(x-12)=0

=(2x²+1)[(4x-3)-(x-12)=0

=(2x²+1)(4x-3-x+12)=0

=(2x²+1)(3x+9)=0

\(\Leftrightarrow\)2x²+1=0 hoặc 3x+9=0

\(\Leftrightarrow\)x=\(\dfrac{1}{2}\)hoặc x=\(\dfrac{-1}{2}\) hoặc x=-3

Vậy S={\(\dfrac{1}{2};\dfrac{-1}{2};-3\)}

e) (2x-1)²+(2-x)(2x-1)=0

=(2x-1)[(2x-1)+(2-x)=0

=(2x-1)(2x-1+2-x)=0

=(2x-1)(x+1)=0

\(\Leftrightarrow\) 2x-1=0 hoặc x+1=0

\(\Leftrightarrow\) x=\(\dfrac{-1}{2}\) hoặc x=-1

Vậy S={\(\dfrac{-1}{2}\);-1}

f)(x+2)(3-4x)=x²+4x+4

=(x+2)(3-4x)=(x+2)²

=(x+2)(3-4x)-(x+2)²=0

=(x+2)[(3-4x)-(x+2)]=0

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

=(x+2)(-5x+1)=0

\(\Leftrightarrow\) x+2=0 hoặc -5x+1=0

\(\Leftrightarrow\) x=-2 hoặc x=\(\dfrac{1}{5}\)

Vậy S={-2;\(\dfrac{1}{5}\)}

a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0

TH2 3x-4=0 <=>x=4/3

KL:.....

b) .

<=> (x+3)(x-1+2x)=0

TH1: x+3=0 <=> x=-3

TH2  x-1=0  <=> x=1

KL:.....

c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 ​\)

KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)

KL:....

a) \(3x^2-4x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)

b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)

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

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

c) \(9x^2+6x+1=0\)

\(\Leftrightarrow\left(3x+1\right)^2=0\)

\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)

d) \(x^2-4x=4\)

\(\Leftrightarrow\left(x-2\right)^2=8\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)

giải đc sao pn dễ mk

chẳng ai giải, thôi mình giải vậy!

a) Đặt \(y=x^2+4x+8\),phương trình có dạng:

\(t^2+3x\cdot t+2x^2=0\)

\(\Leftrightarrow t^2+xt+2xt+2x^2=0\)

\(\Leftrightarrow t\left(t+x\right)+2x\left(t+x\right)=0\)

\(\Leftrightarrow\left(2x+t\right)\left(t+x\right)=0\)

\(\Leftrightarrow\left(2x+x^2+4x+8\right)\left(x^2+4x+8+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}\)vậy tập nghiệm của phương trình là:S={-2;-4}

b) nhân 2 vế của phương trình với 12 ta được:

\(\left(6x+7\right)^2\left(6x+8\right)\left(6x+6\right)=72\)

Đặt y=6x+7, ta được:\(y^2\left(y+1\right)\left(y-1\right)=72\)

giải tiếp ra ta sẽ được S={-2/3;-5/3}

c) \(\left(x-2\right)^4+\left(x-6\right)^4=82\)

S={3;5}

d)s={1}

e) S={1;-2;-1/2}

f) phương trình vô nghiệm

b) \(\dfrac{x-5}{2017}-1+\dfrac{x-2}{2020}-1=\dfrac{x-6}{2016}-1+\dfrac{x-68}{1954}-1\)

​\(\dfrac{x-2022}{2017}+\dfrac{x-2002}{2020}=\dfrac{x-2022}{2016}+\dfrac{x-2022}{1954}\)​

\(\Leftrightarrow\left(x-2022\right)\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\right)=0\)

\(\Leftrightarrow x-2022=0\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\ne0\right)\)

\(\Leftrightarrow x=2022\)