không mất tổng quát ta giả sử 

\(a>b\)

ta có hai trường hợp 1: \(\hept{\begin{cases}x+a>0\\x+b>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-a\\x>-b\end{cases}\Leftrightarrow}}x>-b\)

trường hợp 2 : \(\hept{\begin{cases}x+a< 0\\x+b< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -a\\x< -b\end{cases}\Leftrightarrow}}x< -a\)

Vậy \(\orbr{\begin{cases}x>-b\\x< -a\end{cases}}\) tổng quát \(\orbr{\begin{cases}x>-min\left(a,b\right)\\x< -max\left(a,b\right)\end{cases}}\)

Ta có : (x + a)(x + b) > 0

TH1 : \(\hept{\begin{cases}x+a>0\\x+b< 0\end{cases}}\Leftrightarrow-a< x< -b\)

TH2 : \(\hept{\begin{cases}x+a< 0\\x+b>0\end{cases}}\Leftrightarrow-b< x< -a\)

Nếu a < b => TH1 loại TH2 đúng

Nếu a > b => TH2 loại TH

Nếu a = b => bất phương trình luôn đúng khi \(x\ne a\)

\(\left(x+2\right)\left(x-3\right)\ge0\)

TH1 : \(\hept{\begin{cases}x+2\ge0\\x-3\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-2\\x\ge3\end{cases}}\Leftrightarrow x\ge3\)

TH2 : \(\hept{\begin{cases}x+2\le0\\x-3\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le-2\\x\le3\end{cases}}\Leftrightarrow x\le-2\)

Vậy bft có tập nghiệm S = { x >= 3 ; x =< -2 } 

\(\left(x+2\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2\ge0\\x-3\ge0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\ge-2\\x\ge3\end{cases}}\)

\(\Leftrightarrow x\ge3\)

Hok tốt!!!!!!

a, \(x^2-6x+9=\left(x-3\right)^2\)

b, \(x^2-12x+36=\left(x-4\right)^2\)

c, \(9x^2-25=\left(3x-5\right)\left(3x+5\right)\)

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

e, \(x^4-8x^2+16=\left(x^2-4\right)^2=\left[\left(x-2\right)\left(x+2\right)\right]^2\)

f, \(x^4-81=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)

g, \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=9\left(1-x\right)\left(x+1\right)\)

h, \(\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2\)

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

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

a) x² - 6x + 9 = (x-3)²

b) x² - 12 + 36 = (x-6)²

c) 9x² - 25 = (3x - 25)(3x + 25)

d) x² - x + 1/4 = (x - 1/2)²

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a;b;c\right)\)bài toán trở thành : \(a+b+c=\frac{1}{13};ab+bc+ca=1\)

Tính \(a^2+b^2+c^2\)

Ta có : \(\left(a+b+c\right)^2=\frac{1}{169}< =>a^2+b^2+c^2=\frac{1}{169}-2=-\frac{337}{169}\)

cái kq âm nên loại giùm mình nhé =) cái bt ấy k có giá trị nào thỏa mãn hết chơnnnn

a, \(x^2-6x+9=4< =>\left(x-3\right)^2=4< =>\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

\(< =>\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b,\(x^2\left(x-3\right)-4\left(x-3\right)=0< =>\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)

\(< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}orx=3}\)

c nhường mấy bn khácccc

a) x^2-6x+9=4.

 x=1, x=5

b) x^2(x-3)-(4X-12)=0

x=-2, x=2, x=3

c) (2x+3)^2-4(x+2)^2=12

x=-19/4

a,\(\left(x-1\right)^2-\left(2x\right)^2=0< =>\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(< =>\left(-x-1\right)\left(3x-1\right)=0< =>\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

b,\(\left(3x-5\right)^2-x\left(3x-5\right)=0< =>\left(3x-5\right)\left(3x-5-x\right)=0\)

\(< =>\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}\)

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

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow x=-1;x=\frac{1}{3}\)

b, \(\left(3x-5\right)^2-x\left(3x-5\right)=0\)

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

a;b tìm nhân tử chung ở mẫu bạn tự làm nhé 

c, \(\frac{3}{1-4x}=\frac{2}{4x+1}-\frac{8+6x}{16x^2-1}\)ĐK : \(x\ne\pm\frac{1}{4}\)

\(\Leftrightarrow-\frac{3}{4x-1}=\frac{2}{4x+1}-\frac{8+6x}{\left(4x-1\right)\left(4x+1\right)}\)

\(\Leftrightarrow-\frac{3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\frac{2\left(4x-1\right)-8-6x}{\left(4x+1\right)\left(4x-1\right)}\)

\(\Rightarrow-12x-3=8x-2-8-6x\Leftrightarrow-14x=-7\Leftrightarrow x=\frac{1}{2}\)

i, \(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{x^2+2x-3}\)ĐK : \(x\ne-3;1\)

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

\(\Rightarrow x^2+x-2-x^2-4x-3=4\Leftrightarrow-3x-5=4\Leftrightarrow x=-3\)(ktm)

Vậy pt vô nghiệm 

g, \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)ĐK : \(x\ne1\)

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

\(\Rightarrow3x^2+x-4=4x-4\Leftrightarrow3x^2-3x=0\Leftrightarrow x=0\left(tm\right);x=1\left(ktm\right)\)

h, \(\frac{3}{5x-1}+\frac{2}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)ĐK : \(x\ne\frac{1}{5};\frac{3}{5}\)

\(\Leftrightarrow\frac{3\left(5x-3\right)-2\left(5x-1\right)}{\left(5x-1\right)\left(5x-3\right)}=\frac{-4}{\left(5x-1\right)\left(5x-3\right)}\)

\(\Rightarrow15x-9-10x+2=-4\Leftrightarrow5x=3\Leftrightarrow x=\frac{3}{5}\)(ktm)

Vậy pt vô nghiệm