Với a + b = 1; ab khác 0, ta có :

\(\frac{a}{b^3-1}+\frac{b}{a^3-1}=\frac{a\left(a^3-1\right)+b\left(b^3-1\right)}{\left(b^3-1\right)\left(a^3-1\right)}=\frac{a^4-a+b^4-b}{a^3b^3-a^3-b^3+1}\)

\(=\frac{\left(a^4+b^4\right)-\left(a+b\right)}{a^3b^3-\left(a^3+b^3\right)+1}=\frac{\left(a^2+b^2\right)^2-2a^2b^2-1}{a^3b^3-\left(a+b\right)^2+3ab\left(a+b\right)+1}\)

 \(=\frac{\left[\left(a+b\right)^2-2ab\right]^2-2a^2b^2-1}{a^3b^3+3ab}=\frac{\left(1-2ab\right)^2-2a^2b^2-1}{ab\left(a^2b^2+3\right)}\)

\(=\frac{1-4ab+4a^2b^2-2a^2b^2-1}{ab\left(a^2b^2+3\right)}=\frac{2a^2b^2-4ab}{ab\left(a^2b^2+3\right)}\)

\(=\frac{2ab\left(ab-2\right)}{ab\left(a^2b^2+3\right)}=\frac{2\left(ab-2\right)}{a^2b^2+3}\)(đpcm)

ĐK : x ≥ 0

x - 3√x + 2 = x - 2√x - √x + 2 = √x( √x - 2 ) - ( √x - 2 ) = ( √x - 2 )( √x - 1 ) 

\(\frac{\frac{n-1}{1}+\frac{n-2}{2}+...+\frac{1}{n-1}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}\)

\(=\frac{\left(\frac{n-2}{2}+1\right)+\left(\frac{n-3}{3}+1\right)+...+\left(\frac{1}{n-1}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}\)

\(=\frac{\frac{n}{2}+\frac{n}{3}+...+\frac{n}{n-1}+1}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}\)

\(=\frac{n\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n-1}+\frac{1}{n}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}}=n\)

x³ - ( a + b + c )x² + ( ab + bc + ca )x = abc

<=> x³ - ax² - bx² - cx² + abx + bcx + cax - abc = 0

<=> x³ - ax² - bx² + abx - cx² + bcx + cax - abc = 0

<=> x ( x² - ax - bx + ab ) - c ( x² - bx - ax + ab ) = 0

<=> ( x - c ) ( x² - ax - bx + ab ) = 0

<=> ( x - c ) [ x ( x - b ) - a ( x - b ) ] = 0

<=> ( x - c ) ( x - a ) ( x - b ) = 0

<=>\(\hept{\begin{cases}x-c=0\\x-a=0\\x-b=0\end{cases}}\) <=> a = b = c = x 

\(\left(x+13\right)^4+\left(x+15\right)^4=16\)

Đặt \(x+14=a\), phương trình trở thành:

\(\left(a-1\right)^4+\left(a+1\right)^4=16\)

\(\Leftrightarrow a^4-4a^3+6a^2-4a+1\)\(+a^4+4a^3+6a^2+4a+1=16\)

\(\Leftrightarrow2a^4+12a^2+2=16\).

\(\Leftrightarrow a^4+6a^2+1=8\)

\(\Leftrightarrow a^4+6a^2-7=0\)

\(\Leftrightarrow\left(a^2-1\right)\left(a^2+7\right)=0\)

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

\(\Leftrightarrow\left(x+13\right)\left(x+15\right)\left[\left(x+14\right)^2+7\right]=0\)

Vì \(\left(x+14\right)^2+7\ge7>0\forall x\)nên:

\(\left(x+13\right)\left(x+15\right)=0:\left[\left(x+14\right)^2+7\right]\)

\(\Leftrightarrow\left(x+13\right)\left(x+15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+13=0\\x+15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-13\\x=-15\end{cases}}\)

Vậy phương trình có tập nghiệm: \(S=\left\{-15;-13\right\}\)

a)xét tg ABC và tg MDC có: BAC=DMC=90, ^C chung 

=>tg ABC đ.dạng vs tg MDC(g.g)

b)xét tg ABC và tg MBI có: CAB=BMI=90, ^B chung

=>tg ABC đ.dạng vs tg MBI(g.g)  =>AB/MB=BC/BI=>AB.BI=BM.BC(đpcm)

a) Xét \(\Delta ABC\)và \(\Delta MDC\)

 Ta có: \(\widehat{BAC}=\widehat{DMC}=90^o\)

\(\widehat{C}\)là góc chung

\(\Rightarrow\Delta ABC~\Delta MDC\left(g-g\right)\)

b) Xét \(\Delta BIM\)và \(\Delta BCA\)

Ta có: \(\widehat{IMB}=\widehat{CAB}=90^o\)

\(\widehat{B}\) là góc chung

\(\Rightarrow\Delta BIM~\Delta BCA\left(g-g\right)\)

\(\Rightarrow\frac{BI}{BC}=\frac{BM}{BA}\)

\(\Rightarrow BI\text{.}BA=BM.BC\)

C H I B D A