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

\(\Rightarrow x^2-2x+1+9-x^2=0\)

\(\Rightarrow2x=10\Rightarrow x=5\)

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

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

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

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

a) \(\left(x-1\right)^2+\left(3-x\right)\left(3+x\right)=0\\ \Leftrightarrow x^2-2x+1+9-x^2=0\\ \Leftrightarrow-2x=-10\\ \Leftrightarrow x=5\)

b) \(\left(x-2\right)^2-\left(2x+1\right)^2=0\\ \Leftrightarrow x^2-4x+4-4x^2-4x-1=0\\ \Leftrightarrow-3x^2-8x+3=0\\ \Leftrightarrow3x^2+8x-3=0\\ \Leftrightarrow\left(3x^2+9x\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

1.

a.\(\Leftrightarrow7x-5x=3+12\)

\(\Leftrightarrow2x=15\Leftrightarrow x=\dfrac{15}{2}\)

b.\(\Leftrightarrow6x-10-7x-7=2\)

\(\Leftrightarrow x=-19\)

c.\(\Leftrightarrow1-3x=4x-3\)

\(\Leftrightarrow7x=2\Leftrightarrow x=\dfrac{2}{7}\)

d.\(\Leftrightarrow8x^2-4x+12x-6-8x^2-8x-2=12\)

\(\Leftrightarrow-2=12\left(voli\right)\)

a) Ta có: \(2x^2+3xy+2y^2\)

\(=2\left(x^2+\dfrac{3}{2}xy+y^2\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{3}{4}y+\dfrac{9}{16}y^2+\dfrac{7}{16}y^2\right)\)

\(=2\left(x+\dfrac{3}{4}y\right)^2+\dfrac{7}{8}y^2\ge0\forall x,y\)(đpcm)

ảnh lỗi r ạ

Lỗi rùi

lưu ảnh rồi gửi vào đây nhé

\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

a)
A=0
\(x\left(x-\dfrac{4}{9}\right)=0\)
x=0 hoặc x-4/9=0
x=0 hoặc x=4/9
b)
A>0
\(x\left(x-\dfrac{4}{9}\right)>0\)
TH1
x>0  và x-4/9 >0
x>0  và x>4/9
TH2
x<0 và x-4/9<0
x<0 và x<4/9
c)
A<0
\(x\left(x-\dfrac{4}{9}\right)< 0\)
TH1
x<0 và x-4/9>0
x<0 và x>4/9
TH2
x>0 và x-4/9 <0
x>0 và x<4/9

\(a.\left[{}\begin{matrix}2x+9=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

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

Câu 1: 

a: x+2=0

nên x=-2

b: (x-3)(2x+8)=0

=>x-3=0 hoặc 2x+8=0

=>x=3 hoặc x=-4

a . 

x + 2 = 0

=> x = 0 - 2 = -2 

b ) .

<=> x - 3 = 0 ; 2x + 8 = 0

= > x = 3 ; x = -8/2 = -4 

c ) .

ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5

1)

a) \(x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy S = {\(-2\)}

b) \(\left(x-3\right)\left(2x+8\right)=0\)

\(\Leftrightarrow x-3=0\) hoặc \(2x+8=0\)

*) \(x-3=0\)

\(\Leftrightarrow x=3\)

*) \(2x+8=0\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\)

Vậy S = \(\left\{-4;3\right\}\)

2) ĐKXĐ:

\(x-5\ne0\Leftrightarrow x\ne5\)

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

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

Vậy x=3/2 hoặc x=-5

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

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

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)

b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)

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

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

Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)

c) \(5x\left(2x-3\right)-6x+9=0\)

\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)

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

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

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)

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

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

b: Ta có: \(3x\left(x-2\right)-7\left(x-2\right)=0\)

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

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

c: Ta có: \(5x\left(2x-3\right)-6x+9=0\)

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

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