\(a.b.c=1\)

\(\Leftrightarrow a=1,b=1,c=1\)

\(a+bc=a+b.c=1+1.1=\left(1+1\right)\left(1+1\right)\)

Vậy \(a+b.c=\left(a+b\right)\left(a+c\right)\).

(đúng không?)

Ta có : \(abc=1\Leftrightarrow a=1;b=1;c=1\)(1)

Ta có : \(a+bc=\left(a+b\right)\left(a+c\right)\Leftrightarrow VT=a^2+ac+ab+bc\)(2)

Thay (1) vào (2) ta được : 

\(1+1=1^2+1+1+1\Leftrightarrow2\ne4\)

GiáTrị của Biểu thức là:

\(\left(-3\right)\sqrt{2}\sqrt{11}\sqrt{g}\sqrt{t}+3\sqrt{2}\sqrt{11}+2\sqrt{3^3}\sqrt{5}\)

Ta có:\(x=\sqrt[3]{15+3\sqrt{22}}+\sqrt[3]{15-3\sqrt{22}}\Rightarrow x^3=\left(\sqrt[3]{15+3\sqrt{22}}\right)^3+\left(\sqrt[3]{15-3\sqrt{22}}\right)^3+3\sqrt[3]{\left(15+3\sqrt{22}\right)\left(15-3\sqrt{22}\right)}\left(\sqrt[3]{15+3\sqrt{22}}+\sqrt[3]{15-3\sqrt{22}}\right)\)\(\Rightarrow x^3=15+3\sqrt{22}+15-3\sqrt{22}+3\sqrt[3]{27}x\Rightarrow x^3=30+9x\Rightarrow x^3-9x+1981==2011\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ca=0\Rightarrow\left(a+c\right)\left(b+c\right)=c^2\)

Vì \(a,b>0\)mà \(\frac{1}{c}=-\left(\frac{1}{a}+\frac{1}{b}\right)< 0\)nên \(c< 0\Rightarrow\sqrt{\left(a+c\right)\left(b+c\right)}=-c\)

\(\Rightarrow2c+2\sqrt{\left(a+c\right)\left(b+c\right)}=0\Rightarrow\left(a+c\right)+2\sqrt{\left(a+c\right)\left(b+c\right)}+\left(b+c\right)=a+b\)

\(\Rightarrow\left(\sqrt{a+c}+\sqrt{b+c}\right)^2=a+b\)---> 2 vế đều dương nên ta lấy căn 2 vế:

\(\sqrt{a+c}+\sqrt{b+c}=\sqrt{a+b}\)

 a) ĐK : y >= 2 

Bình phương hai vế 

pt <=> y² + 4 = y² + 4y + 4

    <=> y² - y² - 4y = 4 - 4

    <=> -4y = 0

    <=> y = 0 ( tm )

b) ĐK : y >= 0 

Bình phương hai vế

pt <=> y = y² 

     <=> y² - y = 0

    <=> y( y - 1 ) = 0

    <=> y = 0 ( tm ) hoặc y = 1 ( tm )

a) ĐK là y >= -2 nhé mình đánh thiếu ... xài ip nên hơi khó nhìn

LEVER

\(\frac{3}{a+b\sqrt{3}}-\frac{2}{a-b\sqrt{3}}=7-20\sqrt{3}\)\(\Leftrightarrow\frac{3\left(a-b\sqrt{3}\right)-2\left(a+b\sqrt{3}\right)}{a^2-3b^2}=7-20\sqrt{3}\)

\(\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=7-20\sqrt{3}\)\(\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=\frac{7-20\sqrt{3}}{49-48}\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=\frac{7-20\sqrt{3}}{7^2-3.4^2}\Leftrightarrow\hept{\begin{cases}a=7\\b=4\end{cases}}\)

Ta có: \(\frac{1+A}{1-A}=\frac{1+\frac{x-y}{x+y}}{1-\frac{x-y}{x+y}}=\frac{\frac{2x}{x+y}}{\frac{2y}{x+y}}=\frac{x}{y}\)

Tương tự: \(\frac{1+B}{1-B}=\frac{y}{z};\frac{1+C}{1-C}=\frac{z}{x}\)

\(\Rightarrow\frac{\left(1+A\right)\left(1+B\right)\left(1+C\right)}{\left(1-A\right)\left(1-B\right)\left(1-C\right)}=1\Rightarrowđpcm\)