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

<=> (x-2) (5x-3) = 0

<=> TH1: x - 2 = 0 

          <=> x=2

       TH2: 5x-3 = 0

           <=> x= 3/5

b) (x-2)^2 - (x-1)(x+3)=1

<=> (x²-4x+2) - (x²+3x-x-3) = 1

<=> x²-4x+4-x²-3x+x+3=1
<=> -4x-3x+x= 1-4-3

<=> -6x=-6

<=> x= 1

x=1 nha

HT

k cho mình nha

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

b, \(\left(x-2\right)^2-\left(x-1\right)\left(x+3\right)=1\)

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

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

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

\(\ge1+1+1-\frac{3}{4}=\frac{9}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{2}\)

à quên tách ra mà quên đoạn sau :v thêm vào tí nhé 

\(S\ge\left(a+\frac{1}{4a}\right)+\left(b+\frac{1}{4b}\right)+\left(c+\frac{1}{4c}\right)+\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)

\(\ge2\sqrt{\frac{a}{4a}}+2\sqrt{\frac{b}{4b}}+2\sqrt{\frac{c}{4c}}+\frac{3}{4}.\frac{9}{a+b+c}-\frac{3}{4}\ge1+1+1+\frac{3}{4}.\frac{9}{\frac{3}{2}}-\frac{3}{4}=\frac{27}{4}\)

tích mình với

ai tích mình

mình tích lại

thanks

Tích mình đi mình tích lại

a, Áp dụng \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)

Áp dụng \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x,y>0\)

Ta có: \(A=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(2+\frac{4}{a+b}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)

Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)

b, Áp dụng \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)

Áp dụng \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\forall x,y,z>0\)

Ta có: \(B=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2+\left(1+\frac{1}{c}\right)^2\ge\frac{\left(3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{3}\ge\frac{\left(3+\frac{9}{a+b+c}\right)^2}{3}\ge\frac{\left(3+6\right)^2}{3}=27\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\)

* Các BĐT phụ bạn tự CM nha! Chúc bạn học tốt

Camon bạn!!! Nhưng bạn đọc sai đề r !! ^.^

Sai gì bạn?

Em làm thử nhé!

Bài 1: \(A=\left[\frac{a^2}{b-1}+4\left(b-1\right)\right]+\left[\frac{b^2}{a-1}+4\left(a-1\right)\right]-4\left(a+b\right)+8\)

Cauchy vào là ra rồi ạ;)

Bài 2: Em chịu

2) Có: \(\sqrt{ab}\le\frac{a+b}{2}=1\); \(\sqrt{a}+\sqrt{b}=\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}\le\sqrt{2\left(a+b\right)}=2\)

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

\(\ge\frac{\left(a+b\right)^2}{\sqrt{a}+\sqrt{b}}\ge=\frac{2^2}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=1\)