\(f\left(x\right)=\left(x+1\right)\left(3-x\right)=3x-x^2+3-x\)

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

\(\Delta'=1^2-1\cdot\left(-1\right)\cdot3=4>0,a=-1< 0\)

\(\Rightarrow f\left(x\right)\) có hai nghiệm phân biệt : \(x_1=-1,x_2=3\)

Khi đó \(f\left(x\right)>0\forall x\in\left(-\infty;-1\right)\cup\left(3;+\infty\right)\)

\(f\left(x\right)< 0\forall x\in\left(-1;3\right)\).

Để đa thức \(C\left(x\right),D\left(x\right)\) có nghiệm thì \(C\left(x\right)=0,D\left(x\right)=0\)

Do đó : \(C\left(x\right)=\left(\dfrac{1}{2}\right)^3-2x=0\)

\(\Rightarrow\dfrac{1}{8}-2x=0\)

\(\Rightarrow2x=\dfrac{1}{8}\)

\(\Rightarrow x=\dfrac{1}{8}:2=\dfrac{1}{16}\)

Vậy \(x=\dfrac{1}{16}\) là nghiệm của đa thức \(C\left(x\right)\)

\(D\left(x\right)=2x^2-5x-7=0\)

\(\Rightarrow2x^2+2x-7x-7=0\)

\(\Rightarrow2x\left(x+1\right)-7\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(2x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{-1;\dfrac{7}{2}\right\}\) là nghiệm của đa thức \(D\left(x\right)\)

\(m=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\dfrac{2^{20}}{2^{12}}=2^8=256\)Vậy \(m=256\)

a)\(\dfrac{2}{5}+\dfrac{3}{5}\cdot x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{3}{5}\cdot x=\dfrac{1}{2}-\dfrac{2}{5}\)

\(\Rightarrow\dfrac{3}{5}\cdot x=\dfrac{1}{10}\)

\(\Rightarrow x=\dfrac{1}{10}:\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)

Vậy \(x=\dfrac{1}{6}\)

b) \(\left|2x-1\right|-0,5=1,5\)

\(\Rightarrow\left|2x-1\right|=1,5+0,5=2\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{3}{2};-\dfrac{1}{2}\right\}\)

a) \(\left(x+2\right)\left(x^2-2x+4\right)+\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4+x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-x+6\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\dfrac{23}{4}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(N\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ge\dfrac{23}{4}>0\left(L\right)\end{matrix}\right.\)

Vậy \(S=\left\{-2\right\}\)

b) \(9x^2-4-\left(3x-2\right)^2=0\)

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

\(\Leftrightarrow\left(3x-2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]=0\)

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

\(\Leftrightarrow\left(3x-2\right)\cdot4=0\)

\(\Leftrightarrow3x-2=0\)

\(\Leftrightarrow x=\dfrac{2}{3}\)

Vậy \(S=\left\{\dfrac{2}{3}\right\}\)

Đề như thế này à ?

\(\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

\(=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+2\sqrt{x}-3\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

Lần sau bạn ghi đề bài cho đúng vào

\(4\dfrac{1}{5}+\dfrac{4}{5}+2\dfrac{2}{7}+3\dfrac{5}{7}=\left(4\dfrac{1}{5}+\dfrac{4}{5}\right)+\left(2\dfrac{2}{7}+3\dfrac{5}{7}\right)\)

\(=4+\dfrac{1}{5}+\dfrac{4}{5}+2+\dfrac{2}{7}+3+\dfrac{5}{7}\)

\(=4+1+\left(2+3\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)

\(=5+5+1=11\)

1) \(B=\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{4-x}=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{x-4}=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{2\sqrt{x}+2}{x-4}\)

Thay \(x=16\) vào biểu thức B ta được :

\(B=\dfrac{2\sqrt{x}+2}{x-4}=\dfrac{2\sqrt{16}+2}{16-4}=\dfrac{2\cdot4+2}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)

2) \(P=\dfrac{B}{A}=\dfrac{\dfrac{2\sqrt{x}+2}{x-4}}{\dfrac{2}{\sqrt{x}+2}}=\dfrac{2\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}+2}{2}=\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\cdot2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

3)

\(P< 1\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 1\)

\(\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-1< 0\)

\(\Rightarrow\dfrac{\sqrt{x}+1-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}< 0\)

\(\Rightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)

\(\Rightarrow\dfrac{3}{\sqrt{x}-2}< 0\)

\(\Rightarrow\sqrt{x}-2< 0\)

\(\Rightarrow\sqrt{x}< 2\)

\(\Rightarrow x< 4\Rightarrow x\in\left\{0;1;2;3\right\}\)

\(\left(-2\right)^3=-8\) bạn ạ chứ không phải là bằng 8 nên n = 3 là không đúng rồi 

\(-8x-10=\left|-8x-10\right|\)

\(\left|-8x-10\right|=-8x-10\) khi \(-8x-10\ge0\Leftrightarrow-8x\ge10\Leftrightarrow x\le-\dfrac{5}{4}\)

PT trở thành :

\(-8x-10=-8x-10\)

\(\Leftrightarrow-8x+8x=-10+10\)

\(\Leftrightarrow0x=0\)

PT có vô số nghiệm

Vậy \(S=\){\(x\in R\)| x\(\le-\dfrac{5}{4}\)}

\(\left|-8x-10\right|=-\left(-8x-10\right)=8x+10\) khi \(-8x-10< 0\Leftrightarrow-8x< 10\Leftrightarrow x>-\dfrac{5}{4}\)

PT trở thành :

\(-8x-10=8x+10\)

\(\Leftrightarrow-8x-8x=10+10\)

\(\Leftrightarrow-16x=20\)

\(\Leftrightarrow4x=-5\)

\(\Leftrightarrow x=-\dfrac{5}{4}\)(KTMĐK)

Vậy \(S=\){\(x\in R\)|x \(\le-\dfrac{5}{4}\)}

=> Đáp án B đúng