Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

1.

\(f\left(x\right)=\frac{x-7}{\left(x-4\right)\left(4x-3\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định tại \(x=\left\{\frac{3}{4};4\right\}\)

\(f\left(x\right)=0\Rightarrow x=7\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}\frac{3}{4}< x< 4\\x>7\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}x< \frac{3}{4}\\4< x< 7\end{matrix}\right.\)

2.

\(f\left(x\right)=\frac{11x+3}{-\left(x-\frac{5}{2}\right)^2-\frac{3}{4}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=-\frac{3}{11}\)

\(f\left(x\right)>0\Rightarrow x< -\frac{3}{11}\)

\(f\left(x\right)< 0\Rightarrow x>-\frac{3}{11}\)

3.

\(f\left(x\right)=\frac{3x-2}{\left(x-1\right)\left(x^2-2x-2\right)}\)

Vậy:

\(f\left(x\right)\) ko xác định khi \(x=\left\{1;1\pm\sqrt{3}\right\}\)

\(f\left(x\right)=0\Rightarrow x=\frac{2}{3}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< 1-\sqrt{3}\\\frac{2}{3}< x< 1\\x>1+\sqrt{3}\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow\left[{}\begin{matrix}1-\sqrt{3}< x< \frac{2}{3}\\1< x< 1+\sqrt{3}\end{matrix}\right.\)

4.

\(f\left(x\right)=\frac{\left(x-2\right)\left(x+6\right)}{\sqrt{6}\left(x+\frac{\sqrt{6}}{4}\right)^2+\frac{8\sqrt{2}-3\sqrt{6}}{8}}\)

Vậy:

\(f\left(x\right)=0\Rightarrow x=\left\{-6;2\right\}\)

\(f\left(x\right)>0\Rightarrow\left[{}\begin{matrix}x< -6\\x>2\end{matrix}\right.\)

\(f\left(x\right)< 0\Rightarrow-6< x< 2\)

\(x-\frac{11x^2-5x+6}{x^2+5x+6}>0\)

\(\Leftrightarrow\frac{x^3-6x^2+11x-6}{x^2+5x+6}>0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)}{\left(x+2\right)\left(x+3\right)}>0\Rightarrow\left[{}\begin{matrix}x>3\\1< x< 2\\-3< x< -2\end{matrix}\right.\)

b/ \(\frac{2-x}{x^3+x^2}-\frac{1-2x}{x^3-3x^2}>0\)

\(\Leftrightarrow\frac{\left(2-x\right)\left(x+1\right)-\left(1-2x\right)\left(x-3\right)}{x^2\left(x+1\right)\left(x-3\right)}>0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-5\right)}{x^2\left(x+1\right)\left(x-3\right)}>0\Rightarrow\left[{}\begin{matrix}x< -1\\x>5\\1< x< 3\end{matrix}\right.\)

c/ \(\left|x^2-x-1\right|\le x-1\)

Với \(x< 1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) BPT vô nghiệm

Với \(x\ge1\) hai vế ko âm, bình phương:

\(\left(x^2-x-1\right)^2\le\left(x-1\right)^2\)

\(\Leftrightarrow\left(x^2-x-1\right)^2-\left(x-1\right)^2\le0\)

\(\Leftrightarrow\left(x^2-2x\right)\left(x^2-2\right)\le0\) \(\Rightarrow\sqrt{2}\le x\le2\)

1,\(pt\Leftrightarrow11x^2-5x+6=x^3+5x^2+6x\)

\(\Leftrightarrow x^3-6x^2+11x-6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)(tm)

2,\(pt\Leftrightarrow\frac{1}{x+1}+\frac{2}{x^2-x+1}=\frac{2x+3}{x^3+1}\)

\(\Leftrightarrow\frac{x^2-x+1+2x+2}{x^3+1}=\frac{2x+3}{x^3+1}\)

\(\Rightarrow x^2-x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

a/ ĐKXĐ: ...

\(\Leftrightarrow2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{2\left(x-1\right)}{x}+3\)

Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)

\(\frac{2}{a}-a=2a^2+3\Leftrightarrow2a^3+a^2+3a-2=0\)

\(\Leftrightarrow\left(2a-1\right)\left(a^2+a+2\right)=0\Leftrightarrow a=\frac{1}{2}\)

\(\Rightarrow\sqrt{\frac{x-1}{x}}=\frac{1}{2}\Leftrightarrow4\left(x-1\right)=x\)

b/ ĐKXĐ: ...

\(\Leftrightarrow3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{3\left(x-1\right)}{2x}+10\)

Đặt \(\sqrt{\frac{x-1}{2x}}=a>0\)

\(\frac{3}{a}+4a=3a^2+10\Leftrightarrow3a^3-4a^2+10a-3=0\)

\(\Leftrightarrow\left(3a-1\right)\left(a^2-a+3\right)=0\Leftrightarrow a=\frac{1}{3}\)

\(\Leftrightarrow\sqrt{\frac{x-1}{2x}}=\frac{1}{3}\Leftrightarrow9\left(x-1\right)=2x\)

c/ ĐKXĐ: ...

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

Đặt \(\sqrt{\frac{3-2x}{x}}=a>0\)

\(\frac{1}{a}+5a=4a^2+5\Leftrightarrow4a^3-5a^2+5a-1=0\)

\(\Leftrightarrow\left(4a-1\right)\left(a^2-a+1\right)=0\Leftrightarrow a=\frac{1}{4}\)

\(\Leftrightarrow\sqrt{\frac{3-2x}{x}}=\frac{1}{4}\Leftrightarrow16\left(3-2x\right)=x\)

d/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)

\(a^2-2a=3\Leftrightarrow a^2-2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=3\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{\frac{x-1}{x}}=3\Leftrightarrow x-1=9x\)

1/ Đặt \(\sqrt[3]{x^2+5x-2}=t\Rightarrow x^2+5x=t^3+2\)

\(t^3+2=2t-2\)

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

\(\Leftrightarrow\left(t+2\right)\left(t^2-2t+2\right)=0\)

\(\Rightarrow t=-2\)

\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\)

\(\Leftrightarrow x^2+5x-2=-8\)

\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

2/ \(\Leftrightarrow2x+11+3\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)

\(\Leftrightarrow\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}=-\sqrt[3]{x+6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x+5=-x-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x=-\frac{11}{2}\end{matrix}\right.\)