Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
Ta có : \(P=\sqrt{2x^2+xy+2y^2}+\sqrt{2y^2+yz+2z^2}+\sqrt{2z^2+xz+2x^2}\)
Xét : \(\sqrt{2x^2+xy+2y^2}=\sqrt{\dfrac{3}{4}.\left(x-y\right)^2+\dfrac{5}{4}.\left(x+y\right)^2}\)
\(\ge\sqrt{\dfrac{5}{4}.\left(x+y\right)^2}=\dfrac{\sqrt{5}}{2}.\left(x+y\right)\)
\(CMTT:\sqrt{2y^2+yz+2z^2}\ge\dfrac{\sqrt{5}}{2}.\left(y+z\right)\)
\(\sqrt{2z^2+xz+2x^2}\ge\dfrac{\sqrt{5}}{2}.\left(x+z\right)\)
Do đó : \(P\ge\dfrac{\sqrt{5}}{2}.\left(x+y+y+z+z+x\right)=\dfrac{2\sqrt{5}.\left(x+y+z\right)}{2}\)
\(\Rightarrow P\ge\sqrt{5}.\left(x+y+z\right)\)
Ta có : BĐT : \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
Mà : \(xy+yz+zx=3\)
\(\Rightarrow\left(x+y+z\right)^2\ge9\)
\(\Leftrightarrow x+y+z\ge3\)
\(\Rightarrow P_{min}=3\sqrt{5}\)
Dấu bằng xảy ra : \(\Leftrightarrow x=y=z=1\)
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4
Đề có vấn đề theo tôi đề như sau :
\(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}.\)
Rheo tôi đề như vậy
mong xem lại đề
a) \(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}+\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{3}{\sqrt{x}+3}\)
\(=\frac{15\sqrt{x}-11+3x+7\sqrt{x}-6-3+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{23\sqrt{x}+3x-20}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
Toán này lớp 8 đúng không ta
\(\sqrt{-x^2+2x+2}=\sqrt{3-\left(x^2-2x+1\right)}\)
= \(\sqrt{3-\left(x-1\right)^2}\le\sqrt{3}\)
Đạt được khi x = 1
Câu còn lại làm tương tự
\(A=\sqrt{\left(1989-x\right)^2}+\sqrt{\left(1990-x\right)^2}=\left|x-1989\right|+\left|1990-x\right|\)
Áp dụng bất đẳng thức : \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\). Dấu "=" xảy ra khi a,b cùng dấu.
\(A=\left|x-1989\right|+\left|1990-x\right|\ge\left|x-1989+1990-x\right|=1\)
\(\Rightarrow A\ge1\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1989\ge0\\1990-x\ge0\end{cases}}\Leftrightarrow1989\le x\le1990\)
Vậy Min A = 1 \(\Leftrightarrow1989\le x\le1990\)