\(M=\left(x^2+y^2\right)^2+x^4+x^2y^2+y^2\)

\(M=1+x^2\left(x^2+y^2\right)+y^2\)

\(M=1+x^2+y^2\)

\(M=1+1=2\)

câu b bạn xem lại đề ạ chắc thiếu mất dấu cộng

\(4\left(3x^2+5x+2\right)=0\Leftrightarrow12x^2+20x+8=0\)

\(\Leftrightarrow12x^2+20x+1=-7\)

thanks bn nhaa, mk k cho bn rr

mk k cho bn rr

a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

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

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

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

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

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

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

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

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

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

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

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

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết

a) 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{19}{24}\)

b) 5(2x-3)+4x(x-2)+2x(3-2x)=0

\(\Leftrightarrow\)10x-15+4x²-8x+6x-4x²=0

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

\(\Leftrightarrow x=\dfrac{15}{8}\)

vậy x=\(\dfrac{15}{8}\)

c)3x(2-x)+2x(x-1)=5x(x+3)

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\\ \Leftrightarrow4x-x^2=5x^2+15x\\ \Leftrightarrow4x-x^2-5x^2-15x=0\\ \)

\(\Leftrightarrow-6x^2-11x=0\\ \Leftrightarrow-x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-11}{6}\end{matrix}\right.\)

Vậy x=0 hoặc x=\(\dfrac{-11}{6}\)

\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\left(5x+3\right).5\left(3x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)

\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)

\(\left(3x-2\right)\left(13-4x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)

còn đâu tự lm lười :_# 

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

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

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

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

a) 3x(4x - 3) - 2x(5 - 6x) = 0

=> 6x² - 9x - 10x + 12x² = 0

=> 18x² - 19x = 0

=> x(18x - 19) = 0

=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)

b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0

=> 10x - 15 + 4x² - 8x + 6x - 4x² = 0

=> 8x - 15 = 0

=> 8x = 15

=> x = 15 : 8 = 15/8

c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)

=> 6x - 3x² + 2x² - 2x = 5x² + 15x

=> 4x - x² - 5x² - 15x = 0

=> -6x² - 11x = 0

=> -x(6x - 11) = 0

=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)

a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)

b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)

\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)

\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)

d) \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)

\(\Leftrightarrow14x^2+18=0\)

Mà \(14x^2+18>0\)nên pt vô nghiệm

a) Ta có: 3x(4x-3)-2x(5-6x)=0

\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)

\(\Leftrightarrow24x^2-19x=0\)

\(\Leftrightarrow x\left(24x-19\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{19}{24}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{19}{24}\right\}\)

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

\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)

\(\Leftrightarrow8x-15=0\)

\(\Leftrightarrow8x=15\)

hay \(x=\frac{15}{8}\)

Vậy: \(x=\frac{15}{8}\)

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

\(\Leftrightarrow6x-3x^2+2x^2-2x=5x^2+15x\)

\(\Leftrightarrow-x^2+4x-5x^2-15x=0\)

\(\Leftrightarrow-6x^2-11x=0\)

\(\Leftrightarrow6x^2+11x=0\)

\(\Leftrightarrow x\left(6x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-11}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{-11}{6}\right\}\)

d) Ta có: \(3x\left(x+1\right)-5x\left(3-x\right)+6\left(x^2+2x+3\right)=0\)

\(\Leftrightarrow3x^2+3x-15x+5x^2+6x^2+12x+18=0\)

\(\Leftrightarrow14x^2+18=0\)

\(\Leftrightarrow14x^2=-18\)

mà \(14x^2\ge0\forall x\)

nên \(x\in\varnothing\)

Vậy: \(x\in\varnothing\)

1) Ta có: \(3x\left(2-5x\right)+35x-14=0\)

\(\Leftrightarrow3x\left(2-5x\right)+7\left(5x-2\right)=0\)

\(\Leftrightarrow-3x\left(5x-2\right)+7\left(5x-2\right)=0\)

\(\Leftrightarrow\left(5x-2\right)\left(-3x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-2=0\\-3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=2\\-3x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{2}{5};\dfrac{7}{3}\right\}\)

2) Ta có: \(4x-6+5x\left(3-2x\right)=0\)

\(\Leftrightarrow2\left(2x-3\right)-5x\left(2x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\5x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{2}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{3}{2};\dfrac{2}{5}\right\}\)