ta có : \(\dfrac{x+6\sqrt{x}+9}{x-9}=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)

CHỨNG MINH : 

Đặt \(x=\sqrt{a-1},x\ge0\)Ta có ; \(a=x^2+1\)

\(\Rightarrow a-3\sqrt{a-1}+2=\left(x^2+1\right)-3x+2=x^2-3x+3=\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}\right)+3-\frac{9}{4}=\left(x-\frac{3}{2}\right)^2+\frac{3}{4}>0\)với mọi x.

Vậy \(a-3\sqrt{a-1}+2>0\left(a\ge1\right)\)(ddpcm)

TÍNH : Mình nghĩ đề bài bị sai, để mình sửa lại nhé ^^

Đề bài : Tính \(\left(1-\sqrt{1990}\right)\left(\sqrt{1991+2\sqrt{1990}}\right)\)

Ta có : \(\left(1-\sqrt{1990}\right)\left(\sqrt{1991+2\sqrt{1990}}\right)=\left(1-\sqrt{1990}\right)\left(\sqrt{1990+2\sqrt{1990}+1}\right)=\left(1-\sqrt{1990}\right)\left(\sqrt{\left(\sqrt{1990}+1\right)^2}\right)=\left(1-\sqrt{1990}\right)\left(1+\sqrt{1990}\right)=1^2-\left(\sqrt{1990}\right)^2=1-1990=-1989\)

Câu 2:

\(A-4=2x+3y\Rightarrow\left(A-4\right)^2=\left(2x+3y\right)^2\)

\(\left(A-4\right)^2\le\left(2^2+3^2\right)\left(x^2+y^2\right)=676\)

\(\Rightarrow-26\le A-4\le26\)

\(\Rightarrow-22\le A\le30\)

\(A_{max}=30\) khi \(\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)

\(A_{min}=-22\) khi \(\left\{{}\begin{matrix}x=-4\\y=-6\end{matrix}\right.\)

\(2x+3y=1\Rightarrow y=\frac{1-2x}{3}\)

Do \(x;y\ge0\Rightarrow0\le x\le\frac{1}{2}\)

\(A=x^2+3\left(\frac{1-2x}{3}\right)^2=x^2+\frac{1}{3}\left(4x^2-4x+1\right)=\frac{7}{3}x^2-\frac{4}{3}x+\frac{1}{3}\)

\(A=\frac{7}{3}\left(x-\frac{2}{7}\right)^2+\frac{1}{7}\ge\frac{1}{7}\)

\(\Rightarrow A_{min}=\frac{1}{7}\) khi \(x=\frac{2}{7};y=\frac{1}{7}\)

Mặt khác \(A=\frac{1}{3}x\left(7x-4\right)+\frac{1}{3}\)

Do \(x\le\frac{1}{2}\Rightarrow7x-4< 0\Rightarrow x\left(7x-4\right)\le0\)

\(\Rightarrow A\le\frac{1}{3}\Rightarrow A_{max}=\frac{1}{3}\) khi \(x=0;y=\frac{1}{3}\)

\(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}-\frac{1}{m+n+p}=0\)

\(\Leftrightarrow\frac{m+n}{mn}+\frac{m+n}{p\left(m+n+p\right)}=0\)

\(\Leftrightarrow\left(m+n\right)\left(\frac{pm+pn+p^2+mn}{mnp\left(m+n+p\right)}\right)=0\)

\(\Leftrightarrow\left(m+n\right)\left(n+p\right)\left(p+m\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}m=-n\\m=-p\\p=-n\end{matrix}\right.\)

Cả 3 TH là như nhau

Ví dụ như TH1: \(\frac{1}{m^{2017}}+\frac{1}{-m^{2017}}+\frac{1}{p^{2017}}=\frac{1}{p^{2017}}\)

\(\frac{1}{m^{2017}-m^{2017}+p^{2017}}=\frac{1}{p^{2017}}\) (đpcm)

Đáp án D

Phương pháp:

Cách giải:

+ Với I = U R  → đường đặc trưng V – A có dạng là đường thẳng đi qua gốc tọa độ.

a)\(\sqrt{x}=4\Leftrightarrow x=4^2\Leftrightarrow x=16\)

b)\(\sqrt{x-2}=3\Leftrightarrow x-2=3^2\Leftrightarrow x=9-2=7\)

c)\(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\Leftrightarrow\dfrac{x}{3}-\dfrac{7}{6}=\dfrac{1}{36}\Leftrightarrow\dfrac{x}{3}=-\dfrac{41}{36}\Leftrightarrow x=-\dfrac{41}{12}\)

d)\(x^2=7vớix< 0\)

\(\Leftrightarrow\left(-x\right)^2=7\Leftrightarrow-x=\sqrt{7}\Leftrightarrow x=-\sqrt{7}\)

e)\(x^2-4=0với>0\)

\(\Leftrightarrow x^2=4\Leftrightarrow x=\sqrt{4}=2\)

f)\(\left(2x+7\sqrt{7}\right)^2=7\)

\(\Leftrightarrow4x^2+\sqrt{5488}+343=7\)

\(\Leftrightarrow4x^2+\sqrt{5488}=-336\)

\(\Leftrightarrow4x^2=28\left(12-\sqrt{7}\right)\Leftrightarrow x^2=\dfrac{28\left(12-\sqrt{7}\right)}{4}=7\left(12-\sqrt{7}\right)\)

\(\Leftrightarrow x=\sqrt{7\left(12-\sqrt{7}\right)}=\sqrt{84-7\sqrt{7}}\)

a) \(\sqrt{x}=4\Rightarrow x=16\)

b) \(\sqrt{x-2}-3\\ \Rightarrow x-2=9\\ \Rightarrow x=11\)

c) \(x^2=7\\ \Rightarrow x=\pm\sqrt{7}\\ Vớix< 0\Rightarrow x=-\sqrt{7}\)

d) \(x^2-4=0\\\Rightarrow x=\pm2\\ Vớix>0\Rightarrow x=2 \)