Đặt A = \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\)

A = \(\left(1+\frac{a+b}{a}\right)\left(1+\frac{a+b}{b}\right)\)(Vì a + b = 1)

A = \(\left(2+\frac{b}{a}\right)\left(2+\frac{a}{b}\right)\)

A = \(4+\frac{2a}{b}+\frac{2b}{a}+1\)

A = \(5+2\left(\frac{a}{b}+\frac{b}{a}\right)\)

Vì a, b dương nên áp dụng BĐT Cô - si cho 2 số dương, ta được :

\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{ab}{ba}}\)

\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}\ge2.1=2\)

\(\Leftrightarrow2\left(\frac{a}{b}+\frac{b}{a}\right)\ge4\)

\(\Leftrightarrow5+2\left(\frac{a}{b}+\frac{b}{a}\right)\ge4+5\)

\(\Leftrightarrow A\ge9\)

Dấu bằng xảy ra \(\Leftrightarrow\)a = b > 0

Vậy \(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge9\)với a, b là các số dương và a + b = 1

Tớ quên. Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}a=b>0\\a+b=1\end{cases}}\)

\(\Leftrightarrow a=b=\frac{1}{2}\)

Đặt A =\(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\)

Vì a + b \(\ne\)0 nên A luôn được xác định.

 Giả sử \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)

\(\Leftrightarrow\frac{\left(a^2+b^2\right)\left(a+b\right)^2}{\left(a+b\right)^2}+\frac{\left(ab+1\right)^2}{\left(a+b\right)^2}-\frac{2\left(a+b\right)^2}{\left(a+b\right)^2}\ge0\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)(vì a + b \(\ne\)0)

\(\Leftrightarrow[\left(a^2+2ab+b^2\right)-2ab]\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)

\(\Leftrightarrow[\left(a+b\right)^2-2ab]\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)

\(\Leftrightarrow\left(a+b\right)^4-2ab\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)

\(\Leftrightarrow\left(a+b\right)^4-\left[2ab\left(a+b\right)^2+2\left(a+b\right)^2\right]+\left(ab+1\right)^2\ge0\)

\(\Leftrightarrow\left[\left(a+b\right)^2\right]^2-2\left(a+b\right)^2\left(ab+1\right)+\left(ab+1\right)^2\ge0\)

\(\left[\left(a+b\right)^2-\left(ab+1\right)^2\right]^2\ge0\)(luôn đúng)

Dấu bằng xảy ra 

\(\Leftrightarrow\hept{\begin{cases}a+b\ne0\\\Leftrightarrow a=b\end{cases}}\Leftrightarrow a=b\left(a,b\ne0\right)\)

Vậy \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge\)2 với a, b là các số thỏa mãn a+b \(\ne\)0

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}a=b\\a+b\ne0\end{cases}\Leftrightarrow a=b}\)(a,b \(\ne\)0)

Vậy \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\) với a, b là các số thỏa mãn \(a+b\ne0\)

ta có a^2+b^2=a^2+2ab+b^2-2ab=(a+b)^2-(a+b)ab=(a+b)(a+b-ab)=2(2-ab)

vì a+b=2=> a=2-b

=> 2(2-ab)=2(2-2b+b^2)=2((b-1)^2+1)

vì (b-1)^2>=0 với mọi b

=> (b-1)^2+1>=1 với mọi b

=> 2((b-1)^2+1)>=2 với mọi b

=> a^2+b^2>=2=> a^2+b^2>=a+b

cháu tôi học ghê thế :))

a) 3x³ - 7x² + 17x - 5

= 3x³ - x² - 6x² + 2x + 15x - 5

= x²( 3x - 1 ) - 2x( 3x - 1 ) + 5( 3x - 1 )

= ( 3x - 1 )( x² - 2x + 5 )

b) Đặt A = a² + ab + b² - 3a - 3b + 3

=> 4A = 4a² + 4ab + 4b² - 12a - 12b + 12

= ( 4a² + 4ab + b² - 12a - 6b + 9 ) + ( 3b² - 6b + 3 )

= ( 2a + b - 3 )² + 3( b - 1 )² ≥ 0 ∀ a, b

hay 4A ≥ 0 => A ≥ 0

Dấu "=" xảy ra <=> a = b = 1

a.

\(3x^3-7x^2+17x-5=3x^3-x^2-6x^2+2x+15x-5\)

\(=\left(3x-1\right)\left[x^2-2x+5\right]\)

b.\(a^2+ab+b^2-3a-3b+3=\left(a-1\right)^2+\left(b-1\right)^2+\left(a-1\right)\left(b-1\right)\)

\(=\left[a-1+\frac{b-1}{2}\right]^2+\frac{3}{4}\left(b-1\right)^2\ge0\)

dấu bằng xảy ra khi \(a-1=b-1=0\Leftrightarrow a=b=1\)

x²+4

=x²+4+4x-4x

=(x²+2.x.2+2²)-4x

=(x+2)²-(2√x)²

=(x+2-2√x)(x+2+2√x)

Trả lời:

a, \(\frac{4x-5}{x-1}=2+\frac{x}{x-1}\left(đkxđ:x\ne1\right)\)

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

\(\Rightarrow4x-5=2x-2+x\)

\(\Leftrightarrow4x-5=3x-2\)

\(\Leftrightarrow4x-3x=-2+5\)

\(\Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{3\right\}\)

b, \(\frac{7}{x+2}=\frac{3}{x-5}\left(đkxđ:x\ne-2;x\ne5\right)\)

\(\Leftrightarrow\frac{7\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-5\right)}\)

\(\Rightarrow7x-35=3x+6\)

\(\Leftrightarrow7x-3x=6+35\)

\(\Leftrightarrow4x=41\)

\(\Leftrightarrow x=\frac{41}{4}\left(tm\right)\)

Vậy \(S=\left\{\frac{41}{4}\right\}\)

c, \(\frac{2x+5}{2x}-\frac{x}{x+5}=0\left(đkxđ:x\ne0;x\ne-5\right)\)

\(\Leftrightarrow\frac{2x+5}{2x}=\frac{x}{x+5}\)

\(\Leftrightarrow\left(2x+5\right)\left(x+5\right)=2x^2\)

\(\Leftrightarrow2x^2+15x+25=2x^2\)

\(\Leftrightarrow2x^2+15x-2x^2=-25\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=\frac{-5}{3}\left(tm\right)\)

Vậy \(S=\left\{\frac{-5}{3}\right\}\)

d, \(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\left(đkxđ:x\ne\frac{4}{11}\right)\)

\(\Leftrightarrow\frac{18\left(12x+1\right)+2\left(10x-4\right)\left(11x-4\right)}{18\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\)

\(\Rightarrow216x+18+\left(20x-8\right)\left(11x-4\right)=220x^2+107x-68\)

\(\Leftrightarrow216x+18+220x^2-168x+32=220x^2+107x-68\)

\(\Leftrightarrow220x^2+48x+50=220x^2+107x-68\)

\(\Leftrightarrow220x^2+48x-220x^2-107x=-68-50\)

\(\Leftrightarrow-59x=-118\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy \(S=\left\{2\right\}\)

ĐKXĐ: \(x\ne\pm2\)

Ta có : \(A=\frac{x+1}{x-2}+\frac{x-1}{x+2}+\frac{x^2+3}{4-x^2}\)

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

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

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

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

\(=\frac{x^2+1}{x^2-4}\)

Vì \(x^2+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

\(\Rightarrow\)Để A không âm thì \(x^2-4>0\)(do \(x\ne\pm2\)nên \(x^2-4\ne0\))

\(\Leftrightarrow x^2>4\)

\(\Leftrightarrow\orbr{\begin{cases}x>2\\x< -2\end{cases}}\)

Vậy để A không âm thì \(x>2\)hoặc \(x< -2\)