1.

\(\Leftrightarrow2.4^x-5.2^x+2\le0\)

Đặt \(2^x=t>0\Rightarrow2.t^2-5t+2\le0\)

\(\Rightarrow\frac{1}{2}\le t\le2\Rightarrow\frac{1}{2}\le2^x\le2\)

\(\Rightarrow-1\le x\le1\)

2.

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(5-x\right)>0\\x\left(5-x\right)< 6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}0< x< 5\\\left[{}\begin{matrix}x< 2\\x>3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}0< x< 2\\3< x< 5\end{matrix}\right.\)

3.

\(\Leftrightarrow1\ge\left(\sqrt{6}+\sqrt{5}\right)^{x-1}.\left(\sqrt{6}+\sqrt{5}\right)^{2x-5}\)

\(\Leftrightarrow\left(\sqrt{6}+\sqrt{5}\right)^{3x-6}\le1\)

\(\Leftrightarrow3x-6\le0\Rightarrow x\le2\)

4.

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\1>3^{-x}.3^{\sqrt{x+2}}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\3^{\sqrt{x+2}-x}< 1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\sqrt{x+2}-x< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\sqrt{x+2}\le x\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x+2< x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2-x-2>0\end{matrix}\right.\) \(\Rightarrow x>2\)

5.

\(\Leftrightarrow\left(\frac{4}{3}\right)^{2x-1}.\left(\frac{4}{3}\right)^{-2x^2+x}\ge1\)

\(\Leftrightarrow\left(\frac{4}{3}\right)^{-2x^2+3x-1}\ge1\)

\(\Leftrightarrow-2x^2+3x-1\ge0\)

\(\Rightarrow\frac{1}{2}\le x\le1\)

6.

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\\frac{1}{2}log_2\left(x+7\right)>log_2\left(x+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\\sqrt{x+7}>x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x+7>x^2+2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x^2+x-6< 0\end{matrix}\right.\) \(\Rightarrow-1< x< 2\)

7.

\(\left(\frac{1}{5}\right)^{x^2-2x}\ge\left(\frac{1}{5}\right)^3\)

\(\Leftrightarrow x^2-2x\le3\)

\(\Leftrightarrow x^2-2x-3\le0\Rightarrow-1\le x\le3\)

\(\Rightarrow x=\left\{1;2;3\right\}\Rightarrow\) có 3 nghiệm nguyên dương

\(4\sqrt{2}x\) ạ

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}-x}+\lim\limits_{x\rightarrow-\infty}\dfrac{3x^3-1-x^3}{\sqrt[3]{\left(3x^3-1\right)^2}+x\sqrt[3]{3x^3-1}+x^2}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}+\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{1}{x^2}}{\dfrac{\sqrt[3]{\left(3x^3-1\right)^2}}{x^2}+\dfrac{x\sqrt[3]{3x^3-1}}{x^2}+\dfrac{x^2}{x^2}}=0\)

b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x-x^2}{\sqrt{x^2+x}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^3-x^3+x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{\left(x^3-x^2\right)^2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}+\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x^2}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{x\sqrt[3]{x^3-x^2}}{x^2}+\dfrac{\sqrt[3]{\left(x^3-x^2\right)^2}}{x^2}}\)

\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-1-2x-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{4x^2-1}+\sqrt[3]{\left(2x+1\right)^2}}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x^{\dfrac{2}{3}}}}{\dfrac{\sqrt[3]{\left(2x-1\right)^2}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{4x^2-1}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{\left(2x+1\right)^2}}{x^{\dfrac{2}{3}}}}=0\)

Check lai ho minh nhe :v

Bài 1:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)

\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)

\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)

Bài 2:

\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)

\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)

\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)

1.

ĐKXĐ: \(x=2\)

Xét \(x=2\), bất phương trình vô nghiệm

\(\Rightarrow\) bất phương trình đã cho vô nghiệm

\(\Rightarrow\) Không tồn tại \(a,b\) thỏa mãn

Đề bài lỗi chăng.

a, ĐK: ​\(\left(x+1\right)\left(x^2+2x-1\right)\ge0\)​

\(x^2+5x+2=4\sqrt{x^3+3x^2+x-1}\)

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

TH1: \(x\ge-1\)

\(pt\Leftrightarrow\left(\sqrt{x^2+2x-1}-\sqrt{x+1}\right)\left(\sqrt{x^2+2x-1}-3\sqrt{x+1}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2x-1}=\sqrt{x+1}\\\sqrt{x^2+2x-1}=3\sqrt{x+1}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-1=x+1\\x^2+2x-1=9x+9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-7x-10=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

TH2: \(x< -1\)

\(pt\Leftrightarrow\left(\sqrt{-x^2-2x+1}-\sqrt{-x-1}\right)\left(\sqrt{-x^2-2x+1}-3\sqrt{-x-1}\right)=0\)

\(\Leftrightarrow...\)

Bài này dài nên ... cho nhanh nha, đoạn sau dễ rồi

bài 2

ta có \(\left(\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\right)^2\)

\(=\left(\sqrt{a}.\sqrt{\frac{8a^2+1}{a}}+\sqrt{b}.\sqrt{\frac{8b^2+1}{b}}+\sqrt{c}.\sqrt{\frac{8c^2+1}{c}}\right)^2\)\(=\left(A\right)\)

Áp dụng bất đẳng thức Bunhiacopxki ta có;

\(\left(A\right)\le\left(a+b+c\right)\left(8a+\frac{1}{a}+8b+\frac{1}{b}+8c+\frac{8}{c}\right)\)

\(=\left(a+b+c\right)\left(9a+9b+9c\right)=9\left(a+b+c\right)^2\)

\(\Rightarrow3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)(đpcm)

Dấu \(=\)xảy ra khi \(a=b=c=1\)

câu 1 dễ mà liên hợp đi x=\(\frac{4}{5}\)

câu hình 

ad bđt svacso

\(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_b}\ge\frac{9}{h_a+2h_b}\)

tt vs mấy cái còn lại rồi dùng S=p.r