khử mẫu bt lấy căn :
a) \(3xy\cdot\sqrt{\dfrac{2}{xy}}\)
b)\(x\cdot\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}\)
c) \(xy\cdot\sqrt{\dfrac{1}{xy}}+x\cdot\sqrt{\dfrac{y}{x}}-y\cdot\sqrt{\dfrac{x}{y}}\)
Rút gọn:
\(A=\left[\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\cdot\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}\right]\cdot\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{xy^3}+\sqrt{x^3y}}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{y}=b\end{matrix}\right.\), ta có:
\(A=\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\times\dfrac{2}{a+b}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\)\(\times\dfrac{a^3+ab^2+a^2b+b^3}{ab^3+a^3b}\)
\(=\left(\dfrac{b+a}{ab}\times\dfrac{2}{a+b}+\dfrac{b^2+a^2}{a^2b^2}\right)\)\(\times\dfrac{a^2\left(a+b\right)+b^2\left(a+b\right)}{ab\left(a^2+b^2\right)}\)
\(=\dfrac{2ab+b^2+a^2}{a^2b^2}\times\dfrac{\left(a+b\right)\left(a^2+b^2\right)}{ab\left(b^2+a^2\right)}\)
\(=\dfrac{\left(a+b\right)^3}{a^3b^3}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{\left(xy\right)^3}}\)
\(\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}\cdot\left(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt[]{xy}}\right)\)
Thu gọn biểu thức trên
Ta có: \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{y}}{x+\sqrt{xy}}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}\right)+\sqrt{y}\left(x-\sqrt{xy}\right)}{x^2-xy}\)
\(=\dfrac{\sqrt{y}\left(x+\sqrt{xy}+x-\sqrt{xy}\right)}{x\left(x-y\right)}=\dfrac{2x\sqrt{y}}{x\left(x-y\right)}\)
\(=\dfrac{2\sqrt{y}}{x-y}=\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(\Rightarrow A=\dfrac{\sqrt{x}+\sqrt{y}-1}{x+\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{2\sqrt{xy}}.\dfrac{2\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\sqrt{x}+\sqrt{y}-1+1}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}}{x}\)
rút gọn:
A=\(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\left(a,b\ge0,a\ne b\right)\)
B=\(\left(\dfrac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\cdot\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\right)\left(x,y\ge0,x\ne y\right)\)
Giúp em giải các hệ phương trình này với
a)\(\begin{cases}x^4+2y^3-x=-\dfrac{1}{4}+3\sqrt{3}\\ y^4+2x^3-y=-\dfrac{1}{4}-3\sqrt{3}\end{cases}\)
b) \(\begin{cases} x+\dfrac{78y}{x^2+y^2}=20\\ y+\dfrac{78x}{x^2+y^2}=15\end{cases}\)
c) \(\begin{cases}\left(1-\dfrac{12}{y+3x}\right)\cdot \sqrt{x}=2\\ \left(1+\dfrac{12}{y+3x}\right)\cdot\sqrt{y}=6 \end{cases}\)
d) \(\begin{cases} \sqrt{x+1}+\sqrt[4]{x-1}-\sqrt{y^4+2}=y\\ x^2+2x(y-1)+y^2-6y+1=0\end{cases}\)
e) \(\begin{cases} \sqrt{4x^2+(4x-9)(x-y)}+\sqrt{xy}=3y\\ 4\sqrt{(x+2)(y+2x)}=3(x+3)\end{cases}\)
Bài 50 (trang 30 SGK Toán 9 Tập 1)
Trục căn thức ở mẫu với giả thiết các biểu thức chữ đều có nghĩa
$\dfrac{5}{\sqrt{10}}$; $\dfrac{5}{2 \sqrt{5}}$ ; $\dfrac{1}{3 \sqrt{20}}$ ; $\dfrac{2 \sqrt{2}+2}{5 \sqrt{2}}$ ;$\dfrac{y+b.\sqrt{y}}{b.\sqrt{y}}$.
\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)
\(\frac{5}{2\sqrt{5}}=\frac{10\sqrt{5}}{20}=\frac{\sqrt{5}}{2}\)
\(\frac{1}{3\sqrt{20}}=\frac{3\sqrt{20}}{180}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)
\(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{10\sqrt{2}\left(\sqrt{2}+1\right)}{50}=\frac{20+10\sqrt{2}}{50}=\frac{10\left(2+\sqrt{2}\right)}{50}=\frac{2+\sqrt{2}}{5}\)
\(\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{y\left(\sqrt{y}+b\right)}{by}=\frac{\sqrt{y}+b}{b}\)
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Nguồn : Bài 50 trang 30 SGK Toán 9 tập 1 - loigiaihay.com
#Ye Chi-Lien
\(\dfrac{5}{\sqrt{10}}=\dfrac{\sqrt{10}}{2}\)
\(\dfrac{5}{2\cdot\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{1}{3\cdot\sqrt{20}}=\dfrac{\sqrt{20}}{60}\) ;
Tính
\(\dfrac{1}{x-y}\cdot\sqrt{x^4\left(x-y\right)^2}\) (x>y)
\(\sqrt{27}\cdot\sqrt{48\cdot\left(2-a\right)^2}\) (a>2)
\(\left(\sqrt{2012}+\sqrt{2011}\right)\cdot\left(\sqrt{2012}+\sqrt{2011}\right)\)
\(\sqrt{\dfrac{64x^2}{49\left(y+1\right)^2}}\) (x<0;y>-1)
\(\sqrt{\dfrac{121x^2}{144\left(y+2\right)}}\left(x>0;y< -2\right)\)
\(\sqrt{\dfrac{676x^3}{169xy^2}}\left(x>0;y< 1\right)\)
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
Bài 1: CMR \(P=\dfrac{a+b}{\sqrt{a\cdot\left(3a+b\right)}+\sqrt{b\cdot\left(3b+a\right)}}>=\dfrac{1}{2}\)
với a, b > 0
Bài 2: cho x, y, z > 0. CMR
\(P=\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{x+z}}+\sqrt{\dfrac{z}{x+y}}>2\)
- Khử mẫu của biểu thức lấy căn ( mình làm rồi nhưng hơi nghi ngờ về kết quả nên muốn kiểm tra lại ) :
a) \(x\sqrt{\dfrac{6}{x}}+\sqrt{\dfrac{2x}{3}}\)
b) \(xy\sqrt{\dfrac{1}{xy}}+x\sqrt{\dfrac{y}{x}}-y^2\sqrt{\dfrac{x}{y}}\)
bạn làm rồi nên mk chỉ viết kq thôi nhé :)
a)\(\dfrac{4\sqrt{6x}}{3}\)
b)\(\left(2-y\right)\sqrt{xy}\)
bài 1: tính
a) \(\sqrt{1,2\cdot27}\) b) \(\sqrt{55\cdot77\cdot35}\)
c) (\(\sqrt{3}-\sqrt{2}\) )\(^2\) d) (3\(\sqrt{2}-1\))*(3\(\sqrt{2}+1\))
e) (\(\sqrt{6}+7\)) (\(\sqrt{3}-\sqrt{2}\)) i) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\)
h) \(\sqrt{\sqrt{2}-1}\cdot\sqrt{\sqrt{2}}+1\)
bài 2: tính
a) \(\sqrt{9}-\sqrt{17}\cdot\sqrt{9}+\sqrt{17}\)
b) 2\(\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
c) \(\dfrac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\) d) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\)
e) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\) f) \(\dfrac{x+\sqrt{xy}}{9+\sqrt{xy}}\) (xy>0)
Bài 1:
a: \(=\sqrt{32.4}=\dfrac{9}{5}\sqrt{10}\)
b: \(=\sqrt{5\cdot5\cdot7\cdot7\cdot11\cdot11}=5\cdot7\cdot11=385\)
c: \(=5-2\sqrt{6}\)
d: \(=18-1=17\)
e: \(=3\sqrt{2}-2\sqrt{3}+7\sqrt{3}-7\sqrt{2}=-4\sqrt{2}+5\sqrt{3}\)