Bài 1 :

a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

\(A=\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)

\(\Leftrightarrow A=\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}:\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(\Leftrightarrow A=\frac{1}{\sqrt{x}+1}:\frac{1}{\sqrt{x}-2}\)

\(\Leftrightarrow A=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)

b) Để \(A< -1\)

\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< -1\)

\(\Leftrightarrow\sqrt{x}-2< -\sqrt{x}-1\)

\(\Leftrightarrow2\sqrt{x}< 1\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{2}\)

\(\Leftrightarrow x< \frac{1}{4}\)

Vậy để \(A< -1\Leftrightarrow x< \frac{1}{4}\)

a, Với \(x\ge0;x\ne1\)

\(B=\frac{1}{\sqrt{x}-1}=2\Rightarrow2\sqrt{x}-2=1\Leftrightarrow2\sqrt{x}-3=0\Leftrightarrow x=\frac{9}{4}\)

b, Ta có : \(A.B=\frac{x+3}{\sqrt{x}+1}.\frac{1}{\sqrt{x}-1}=\frac{x+3}{x-1}=\frac{x-1+4}{x-1}=1+\frac{4}{x-1}\)

\(\Rightarrow x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

c, Ta có : \(A=\frac{x+3}{\sqrt{x}+1}\le3\Leftrightarrow\frac{x+3}{\sqrt{x}+1}-3\le0\)

\(\Leftrightarrow\frac{x-3\sqrt{x}}{\sqrt{x}+1}\le0\Rightarrow\sqrt{x}-3\le0\Leftrightarrow x\le9\)

Kết hợp với đk vậy 0 =< x =< 9 

x-3 ; mình đánh thiếu

ĐKXĐ: \(x\ge\frac{1}{4}\)

\(\sqrt{5x+1}\le3\sqrt{x}+\sqrt{4x-1}\)

\(\Leftrightarrow5x+1\le9x+4x-1+6\sqrt{4x^2-x}\)

\(\Leftrightarrow3\sqrt{4x^2-x}\ge1-4x\)

Do \(x\ge1\Rightarrow\left\{{}\begin{matrix}1-4x\le0\\\sqrt{4x^2-x}\ge0\end{matrix}\right.\) \(\Rightarrow\) BPT luôn đúng

Vậy nghiệm của BPT là \(x\ge\frac{1}{4}\)

b/ ĐKXĐ: \(x\ge4\)

\(\Leftrightarrow\sqrt{2\left(x^2-16\right)}+x-3>7-x\)

\(\Leftrightarrow\sqrt{2\left(x^2-16\right)}>10-2x\)

- Với \(x>5\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT luôn đúng

- Với \(x\le5\) bình phương 2 vế:

\(2\left(x^2-16\right)>4\left(x-5\right)^2\)

\(\Leftrightarrow x^2-20x+66< 0\)

\(\Rightarrow10-\sqrt{34}< x< 10+\sqrt{34}\)

Vậy nghiệm của BPT là \(x>10-\sqrt{34}\)

ĐKXĐ: \(x>0\)

\(\Leftrightarrow2+\sqrt{x}\le6\sqrt{x}\)

\(\Leftrightarrow5\sqrt{x}\ge2\)

\(\Leftrightarrow\sqrt{x}\ge\frac{2}{5}\)

\(\Rightarrow x\ge\frac{4}{25}\)

a) \(x+3+\sqrt{x^2-6x+9}\left(x\le3\right)\)

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

\(=x+3+\left|x-3\right|\)

\(=x+3-\left(x-3\right)\)

\(=x+3-x+3\)

\(=6\)

b) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\left(-2\le x\le0\right)\)

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

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

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

\(=x+2+x\)

\(=2x+2=2\left(x+1\right)\)

c) \(\frac{\sqrt{x^2-2x+1}}{x-1}\left(x>1\right)\)

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

\(=\frac{\left|x-1\right|}{x-1}\)

\(=\frac{x-1}{x-1}=1\)

d) \(\left|x-2\right|+\frac{\sqrt{x^2-4x+4}}{x-2}\)

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

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

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

\(=\left|x-2\right|-1\)

\(=-\left(x-2\right)-1\)

\(=-x+2-1\)

\(=-x+1=-\left(x-1\right)\)

a) \(\sqrt{4-5x}=12\)

ĐK : x ≤ 4/5

Bình phương hai vế

⇔ \(4-5x=144\)

⇔ \(-5x=140\)

⇔ \(x=-28\)( tm )

b) \(\sqrt{1-4x+4x^2}=5\)

⇔ \(\sqrt{\left(1-2x\right)^2}=5\)

⇔ \(\left|1-2x\right|=5\)

⇔ \(\orbr{\begin{cases}1-2x=5\\1-2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)

ĐK : x ≥ -5

⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)

⇔ \(\left|2\right|\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot\left|3\right|\sqrt{x+5}=6\)

⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)

⇔ \(\frac{5}{4}\sqrt{x+5}=6\)

⇔ \(\sqrt{x+5}=\frac{24}{5}\)

⇔ \(x+5=\frac{576}{25}\)

⇔ \(x=\frac{451}{25}\)( tm )

d)\(\sqrt{x-2}\le3\)

ĐK : x ≥ 2

⇔ \(x-2\le9\)

⇔ \(x\le11\)

Kết hợp với điều kiện => Nghiệm của bpt là 2 ≤ x ≤ 11

Áp dụng bất đẳng thức Bunhia ta có :

\(\left(\sqrt{1+x^2}+\sqrt{2x}\right)^2\le2\left(1+x^2+2x\right)=2\left(x+1\right)^2\text{ nên }\sqrt{1+x^2}+\sqrt{2x}\le\sqrt{2}\left(x+1\right)\)

tương tự ta có : \(\hept{\begin{cases}\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\\\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\end{cases}}\)

Nên \(A\le\sqrt{2}\left(x+y+z+3\right)+\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\left(2-\sqrt{2}\right)\)

\(\le6\sqrt{2}+\left(2-\sqrt{2}\right)\sqrt{3\left(x+y+z\right)}\le6\sqrt{2}+\left(2-\sqrt{2}\right).3=6+3\sqrt{2}\)

dấu bằng xảy ra khi x=y=z=1

ủa bạn oi nó là \(\sqrt{2}x\)mà có phai\(\sqrt{2x}dau\)

Ta sẽ chứng minh

\(\sqrt{x^2+1}+2\sqrt{x}\le\frac{2+\sqrt{2}}{2}\left(x+1\right)\)

\(\Leftrightarrow\left(\sqrt{x^2+1}+2\sqrt{x}\right)^2\le\frac{3+2\sqrt{2}}{2}\left(x+1\right)^2\)

\(\Leftrightarrow\frac{1+2\sqrt{2}}{2}\left(x^2+1\right)-4\sqrt{x\left(x^2+1\right)}+\left(2\sqrt{2}-1\right)x\ge0\)

\(\Leftrightarrow\left(\sqrt{x^2+1}-\sqrt{2x}\right)\left(\frac{1+2\sqrt{2}}{2}\sqrt{x^2+1}-\frac{4-\sqrt{2}}{2}\sqrt{x}\right)\ge0\)

BĐT trên luôn đúng do \(x^2+1\ge2x\)

Vậy ta có:\(\text{∑}\sqrt{x^2+1}+2\sqrt{x}\le\text{∑}\frac{2+\sqrt{2}}{2}\left(x+1\right)\le6+3\sqrt{2}\)

Đẳng thức xảy ra khi x=y=z=1

tích trước trả lời sau

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