a)\(\left(a-2b\right)^2+\left(2a-b\right)^2\ge a^2+b^2\Leftrightarrow\left(a-2b\right)^2-b^2+\left(2a-b\right)^2-a^2\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a-3b\right)+\left(a-b\right)\left(3a-b\right)\ge0\Leftrightarrow\left(a-b\right)\left(4a-4b\right)\ge0\Leftrightarrow4\left(a-b\right)^2\ge0\)(luôn đúng)

Dấu = xảy ra khi a=b

b) \(a^2+b^2+c^2+\frac{3}{4}\ge a+b+c\Leftrightarrow4a^2+4b^2+4c^2+3\ge4a+4b+4c\)

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

\(\Leftrightarrow\left(2a-1\right)^2+\left(2b-1\right)^2+\left(2c-1\right)^2\ge0\)(luôn đúng)

Dấu = xảy ra khi a=b=c=1/2

c)\(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)

Dấu = xảy ra khi a=b=c

Cho các số dương x,y,z t/m 1/x+1/y+1/z=4.

Cm: 1/(2x+y+z)+1/(x+2y+z)+1/(x+y+2z) <=1

\(\left(x^2-4\right)-\left(4x^2+4x+1\right)-2x+3x^2=0\)

\(\Leftrightarrow\left(x^2+3x^2-4x^2\right)+\left(-4x-2x\right)+\left(-4-1\right)=0\)

\(\Leftrightarrow-6x-5=0\Leftrightarrow x=-\frac{5}{6}\)

Vậy nghiệm phương trình là \(x=-\frac{5}{6}\)

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

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

\(\Leftrightarrow x^2-4-4x^2-4x-1-2x+3x^2=0\)

\(\Leftrightarrow-5-6x=0\)

\(\Leftrightarrow-6x=5\Leftrightarrow x=\frac{-5}{6}\)

a, Ta có: \(a^2+ab+b^2=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge0\forall a,b\)

\(\Rightarrow a^2+ab+b^2\ge0\left(đpcm\right)\)

a) Ta có: \(a^2+ab+b^2=a^2+2a\frac{b}{2}+\frac{b^2}{4}+\frac{3b^2}{4}=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge0\)\(\ge0\)

b) \(\left(a+2b\right)^2\ge8ab\Leftrightarrow a^2-4ab+4b^2\ge0\Leftrightarrow\left(a-2b\right)^2\ge0\)(luôn đúng)

\(\left(x-1\right)^3-x\left(x-1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)

\(< =>\left(x-1+x\right)\left(x-1\right)^2=10x-5x^2-11x-22\)

\(< =>-x^2+x-1-10x+5x^2+11x+22=0\)

\(< =>4x^2+3x+21=0\)

\(< =>\left(2x\right)^2+2.2x.\frac{3}{4}+\left(\frac{3}{4}\right)^2+20\frac{9}{25}=0\)

\(< =>\left(2x+\frac{3}{4}\right)^2+20\frac{9}{25}=0\)

Do \(\left(2x+\frac{3}{4}\right)^2\ge0=>\left(2x+\frac{3}{4}\right)^2+20\frac{9}{25}\ge20\frac{9}{25}>0\)

Vậy phương trình vô nghiệm

Dòng 2 là (x-1-x) nha @@

Áp dụng BĐT Cauchy - schwarz dạng Engel:

\(a^2+b^2=\frac{a^2}{1}+\frac{b^2}{1}\ge\frac{\left(a+b\right)^2}{1+1}=\frac{\left(a+b\right)^2}{2}\)

Ta có: \(\left(a-b\right)^2\ge0\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow a^2+2ab+b^2\ge4ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow\frac{\left(a+b\right)^2}{2}\ge2ab\)

Vậy \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\ge2ab\)

Áp dụng bđt bunhiacopxki ta có:

\(\frac{a^2}{1}+\frac{b^2}{1}\ge\frac{\left(a+b\right)^2}{1+1}=\frac{\left(a+b\right)^2}{2}\)(1)

Ta có:\(\frac{\left(a+b\right)^2}{2}\ge2ab\Leftrightarrow\left(a+b\right)^2\ge4ab\)

                                           \(\Leftrightarrow a^2+b^2\ge0\)(luôn đúng vì:\(a^2\ge0,b^2\ge0\forall a,b\inℝ\))

                                           \(\frac{\Rightarrow\left(a+b\right)^2}{2}\ge2ab\)(2)

Từ (1) và (2) suy ra đề bài cần chứng minh

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

\(\Rightarrow x\left(x^2+6x+9\right)-3x=x^3+6x^2+12x+8+1\)

\(\Rightarrow x^3+6x^2+9x-3x=x^3+6x^2+12x+9\)

\(\Rightarrow9x-3x=12x+9\)

\(\Rightarrow6x=12x+9\Rightarrow-6x=9\Rightarrow x=\frac{-3}{2}\)

\(PT\Leftrightarrow x\left(x^2+6x+9\right)-3x=x^3+6x^2+12x+9.\)

\(\Leftrightarrow x^3+6x^2+6x=x^3+6x^2+12x+9\)

\(\Leftrightarrow6x=-9\Rightarrow x=\frac{-3}{2}\)

Ta có: \(\left(a+\frac{1}{2}\right)^2\ge0\forall a\inℝ\)

\(\Rightarrow a^2+a+\frac{1}{4}\ge0\forall a\inℝ\)

\(\Rightarrow a^2+a\ge\frac{-1}{4}\forall a\inℝ\)

(Dấu "="\(\Leftrightarrow a=\frac{-1}{2}\))

Vậy \(a^2+a\ge\frac{-1}{4}\left(đpcm\right)\)

8.3^x+3.2^x-6^x=24

=>6^x-8.3^x-(3.2^x-24)=0

=>3^x(2^x-8)-3(2^x-8)=0

=>(2^x-8)(3^x-3)=0

=>\(\orbr{\begin{cases}2^x-8=0\\3^x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}2^x=8\\3^x=3\end{cases}\Rightarrow}\orbr{\begin{cases}2^x=2^3\\3^x=3^1\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy x=3 hoặc x=1