cănx<4

<=>x>=0 và x²<8

<=>x>=o và -2 căn 2<x<2 căn2

<=>0<=x<2 căn 2

\(\sqrt{x}< 4\)

\(\Rightarrow0< x< 16\)

\(x^2-5\)

\(=\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)\)

tự nhiên quên cách làm  các bạn giúp mik với

9. \(P=5+17^{2019}+2020^{2021}\)

Có \(17\equiv1\left(mod4\right)\Leftrightarrow17^{2019}\equiv1\left(mod4\right)\)

\(2020\equiv0\left(mod4\right)\Leftrightarrow2020^{2021}\equiv0\left(mod4\right)\)

Do đó \(P\equiv5+1\left(mod4\right)\equiv2\left(mod4\right)\)

Ta có đpcm. 

8. \(p>3\)nên \(p=3k+1\)hoặc \(p=3k+2\)

- \(p=3k+1\): \(A=2p^2+3p+4=2\left(3p+1\right)^2+3\left(3p+1\right)+4\equiv2+4\left(mod3\right)\equiv0\left(mod3\right)\)

- \(p=3k+2\): \(A=2p^2+3p+4=2\left(3p+2\right)^2+3\left(3p+2\right)+4\equiv2.2^2+4\left(mod3\right)\equiv0\left(mod3\right)\)

Do đó ta có đpcm. 

Tớ chứng minh câu b trước

Ta có BĐT luôn đúng: \(\left(ay-bx\right)^2\ge0\)

Mà \(\left(ay-bx\right)^2\ge0\Leftrightarrow a^2y^2-2aybx+b^2x^2\ge0\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2-a^2x^2-2axby-b^2y^2\ge0\)

\(\Leftrightarrow a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)-\left(ax+by\right)^2\ge0\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\left(đpcm\right)\)

a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd\)\(+b^2c^2\)

\(=a^2c^2+a^2d^2+b^2d^2+b^2c^2\)\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)

\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)(đpcm)

\(B=\sqrt{x-2}+\sqrt{11-x}\)

\(=\sqrt{x-2}.1+\sqrt{11-x}.1\)

\(\le\sqrt{\left|\left(x-2+11-x\right)\left(1+1\right)\right|}=3\sqrt{2}\)

Dấu \(=\)khi \(x-2=11-x\Leftrightarrow x=\frac{13}{2}\).

ĐKXĐ:\(2\le x\le11\)

Có:\(B=\sqrt{x-2}+\sqrt{11-x}\)

Áp dụng bđt B.C.S ta có:

\(|\sqrt{x-2}+\sqrt{11-x}|\le\sqrt{\left(1+1\right)\left(x-2+11-x\right)}\)

\(\Leftrightarrow|\sqrt{x-2}+\sqrt{11-x}|\le3\sqrt{2}\)

Mà\(\sqrt{x-2}+\sqrt{11-x}\le\left|\sqrt{x-2}+\sqrt{11-x}\right|\)

\(\sqrt{x-2}+\sqrt{11-x}\le3\sqrt{2}\)

Hay\(B\le3\sqrt{2}\)

Dấu "=" xảy ra <=>\(x-2=11-x\)

\(\Leftrightarrow x=\frac{13}{2}\)(Thỏa mãn ĐKXĐ)

Vậy Max\(B=3\sqrt{2}\)\(\Leftrightarrow x=\frac{13}{2}\)

\(-2\sqrt{x^2}+2x+8\)

\(=-2\left|x\right|+2x+8\)

\(=-2x+2x+8\)

\(=8\)