a) Ta có BĐT luôn đúng \(\left(a-b\right)^2\ge0\)\(\Leftrightarrow a^2-2ab+b^2\ge0\)\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow2a^2+2b^2\ge a^2+2ab+b^2\)\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)(đpcm)

b) Ta có các BĐT luôn đúng \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}}\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

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

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)

\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

Answer:

Có: 

\(\left(a-b\right)^2\ge0\)

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

\(\Leftrightarrow2a^2+2b^2\ge a^2+b^2+2ab\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

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

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

\(\Leftrightarrow\)\(2a² + 2c² + 2b² ≥ 2ab + 2ac + 2bc\)

\(\Leftrightarrow\)\(3a² + 3b² + 3c² ≥ a² + b² + c² + 2ab + 2ac + 2bc\)

\(\Leftrightarrow\)\(3(a² + b² + c²) ≥ (a+b+c)²\)

a, tự vẽ nhé 

b, d // d' <=> \(\hept{\begin{cases}2m-3=3\\1\ne-2\left(luondung\right)\end{cases}}\Leftrightarrow m=3\)

c, d vuông d'' <=> \(3a'=-1\Leftrightarrow a'=-\frac{1}{3}\)

\(2020-\sqrt{x^2-2x+1}=1\Leftrightarrow2020-\sqrt{\left(x-1\right)^2}=1\)

\(\Leftrightarrow2020-\left|x-1\right|=1\Leftrightarrow\left|x-1\right|=2019\)

TH1 : \(x-1=2019\Leftrightarrow x=2020\)

TH2 : \(x-1=-2019\Leftrightarrow x=-2018\)

456-345=111

TL:

456 -345 =111

Ht

a) Xét ΔEAM và ΔNAD có 

AE=AN(gt)

ˆEAM=ˆNADEAM^=NAD^(hai góc đối đỉnh)

AM=AD(A là trung điểm của MD)

Do đó: ΔEAM=ΔNAD(c-g-c)

Suy ra: ME=ND(Hai cạnh tương ứng)

ứdfrthyjuiopoikujyhgtf

Ta có:

\(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)

\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}.\text{ Vì thế, }A=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-...-\frac{1}{\sqrt{401}}< 1.\)

chịu rồi

\(\frac{1}{\sqrt{2}}\)+ \(\frac{1}{\sqrt{3}}\)+ \(\frac{1}{\sqrt{4}}\)+ ......... + \(\frac{1}{\sqrt{400}}\)\(< 38\)

Ta chứng minh \(\frac{1}{k}\)\(< \frac{2}{\sqrt{k}\sqrt{k-1}}\)với mọi với mọi \(k\text{∈}N\cdot,k>2\)

Gỉa sử

\(\frac{1}{k}< \frac{2}{\sqrt{k}\sqrt{k-1}}\)\(k\text{∈}N\cdot,k>2\)

\(=\sqrt{k}+\sqrt{k-1}< 2\sqrt{k}=\sqrt{k-1}< \sqrt{k}< k-1< k\)

Khi đó ta có :

\(\frac{1}{\sqrt{k}}\)\(< \frac{2}{\sqrt{k}\sqrt{k-1}}\)\(< \frac{2\left(\sqrt{k}\sqrt{k-1}\right)}{k-\left(k-1\right)}\)\(=2\left(\sqrt{k}\sqrt{k-1}\right)\)

\(VT\left(\cdot\right)< 2\left(\sqrt{2}+\sqrt{1}+\sqrt{3}-\sqrt{2}+......+\sqrt{400}-\sqrt{399}\right)\)

\(VT\left(\cdot\right)< 2\left(\sqrt{400}-1\right)=2.\left(20-1\right)=38\left(dpcm\right)\)

tu di ma lam lop 9 r