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26 tháng 9 2018
\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)
8 tháng 10 2022
a: \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}\)
\(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2\left(x+2\sqrt{xy}+y\right)}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
28 tháng 5 2022
a: \(=\dfrac{2\sqrt{7}-10-6+\sqrt{7}}{4}+\dfrac{24+6\sqrt{7}-20+5\sqrt{7}}{9}\)
\(=\dfrac{3\sqrt{7}-16}{4}+\dfrac{4+11\sqrt{7}}{9}\)
\(=\dfrac{27\sqrt{7}-144+16+44\sqrt{7}}{36}=\dfrac{71\sqrt{7}-128}{36}\)
b: \(=\dfrac{\sqrt{y}\left(x+y\right)}{\sqrt{xy}}\cdot\dfrac{\sqrt{x}-\sqrt{y}}{x+y}\)
\(=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}\)
c: \(=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)+3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right)\cdot\dfrac{3\sqrt{x}-1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+3\sqrt{x}-1}{3\sqrt{x}+1}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)}\cdot\dfrac{1}{3\sqrt{x}-5}\)
\(=\dfrac{3x+\sqrt{x}-2}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-5\right)}\)
12 tháng 10 2022
a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
HS
1 tháng 8 2018
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
HS
1 tháng 8 2018
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 12 2018
ĐKXĐ: \(x\ge3;y\ge1\)
\(\sqrt{x-3}-\sqrt{y-1}+\sqrt[3]{x^2+x+1}-\sqrt[3]{y^2+5y+7}=0\)
\(\Leftrightarrow\dfrac{x-y-2}{\sqrt{x-3}+\sqrt{y-1}}+\dfrac{x^2+x+1-y^2-5y-7}{\sqrt[3]{\left(x^2+x+1\right)}+\sqrt[3]{\left(x^2+x+1\right)\left(y^2+5y+7\right)}+\sqrt[3]{y^2+5y+7}}=0\)
Để cho gọn gàng, ta đặt:
\(\left\{{}\begin{matrix}\sqrt[3]{\left(x^2+x+1\right)}+\sqrt[3]{\left(x^2+x+1\right)\left(y^2+5y+7\right)}+\sqrt[3]{y^2+5y+7}=b>0\\\sqrt{x-3}+\sqrt{y-1}=a>0\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x-y-2}{a}+\dfrac{x^2-y^2-4y-4+x-y-2}{b}=0\)
\(\Leftrightarrow\dfrac{x-y-2}{a}+\dfrac{x^2-\left(y+2\right)^2+\left(x-y-2\right)}{b}=0\)
\(\Leftrightarrow\dfrac{x-y-2}{a}+\dfrac{\left(x-y-2\right)\left(x+y+3\right)}{b}=0\)
\(\Leftrightarrow\left(x-y-2\right)\left(\dfrac{1}{a}+\dfrac{x+y+3}{b}\right)=0\)
\(\Leftrightarrow x-y-2=0\) do \(\left\{{}\begin{matrix}x\ge3\\y\ge1\end{matrix}\right.\) \(\Rightarrow x+y+3>0\Rightarrow\dfrac{1}{a}+\dfrac{x+y+3}{b}>0\)
\(\Rightarrow x=y+2\)
Thay vào Q ta được:
\(Q=y^2-\left(y+2\right)^2+3\left(y+2\right)+4\sqrt{y}+4\)
\(\Rightarrow Q=-y+4\sqrt{y}+6=10-\left(y-4\sqrt{y}+4\right)=10-\left(\sqrt{y}-2\right)^2\le10\)
\(\Rightarrow Q_{max}=10\) khi \(\sqrt{y}-2=0\Rightarrow\left\{{}\begin{matrix}y=4\\x=6\end{matrix}\right.\)
AN
4 tháng 10 2019
f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
NT
2
TL
13 tháng 7 2017
ĐKXĐ \(-1\le x,y\le7\)
TA THẤY: \(\sqrt{x+1}+\sqrt{7-y}=\sqrt{y+1}+\sqrt{7-x}\)
DÙNG PHƯƠNG PHÁP ĐÁNH GIÁ :
NẾU \(x>y\) THÌ \(\hept{\begin{cases}\sqrt{x+1}>\sqrt{y+1}\\\sqrt{7-y}>\sqrt{7-x}\end{cases}}\)DO ĐÓ: \(\sqrt{x+1}+\sqrt{7-y}>\sqrt{y+1}+\sqrt{7-x}\)(VÔ LÍ)
NẾU \(x< y\)THÌ \(\hept{\begin{cases}\sqrt{x+1}< \sqrt{y+1}\\\sqrt{7-y}< \sqrt{7-x}\end{cases}}\)DO ĐÓ \(\sqrt{x+1}+\sqrt{7-y}< \sqrt{y+1}+\sqrt{7-x}\)(VÔ LÍ)
VẬY \(x=y\)THAY VÀO PT(1) TA ĐƯỢC:
\(\sqrt{x+1}+\sqrt{7-x}=4\)
\(\Rightarrow x+1+7-x+2\sqrt{\left(x+1\right)\left(7-x\right)}=16\)
\(\Leftrightarrow\sqrt{-x^2+6x+7}=4\)
\(\Rightarrow-x^2+6x+7=16\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow x=3\)(THỎA MÃN ĐKXĐ)
VẬY HỆ PT CÓ NGHIỆM \(\left(x;y\right)\)LÀ \(\left(3;3\right)\)
12 tháng 11 2021
5: \(=\dfrac{1}{x-y}\cdot x^3\cdot\left(x-y\right)^2=x^3\left(x-y\right)\)
ĐKXĐ ....
x=7-y
Thay vào biểu thức ở trên rồi bình phương hai vế lên