\(\frac{8\sqrt{x}-x-31}{-x+8\sqrt{x}-15}-\frac{\sqrt{x}+5}{\sqrt{x}-3}-\frac{3\sqrt{x}-1}{5-\sqrt{x}}\)
Rút gọn?
rút gọn biểu thức sau với \(x\ge0\),\(x\ne9\),\(x\ne25\)
\(\frac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\frac{\sqrt{x}+5}{\sqrt{x}-3}-\frac{3\sqrt{x}-1}{5-\sqrt{x}}\)
\(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}-15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\) rút gọn giùm ae
Sửa đề: \(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)
Ta có: \(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)
\(=\dfrac{-x+8\sqrt{x}-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}-\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{-x+8\sqrt{x}-31-\left(x-25\right)+3x-9\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{2x-2\sqrt{x}-28-x+25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-3\sqrt{x}+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
\(P=\frac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\frac{\sqrt{x}+15}{\sqrt{x}-3}-\frac{3\sqrt{x}-1}{5-\sqrt{x}}\)
a. Rút gọn P
b. Tìm x sao cho P<1
c.
tìm \(x\in Z\)để \(P\in Z\)
Giúp mk nha mn! Mk đang cần gấp!
\(a,\)\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne9;x\ne25\end{cases}}\)
\(P=\frac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}\)\(-\frac{\sqrt{x}+15}{\sqrt{x}-3}-\frac{3\sqrt{x}-1}{5-\sqrt{x}}\)
\(=\frac{8\sqrt{x}-x-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)\(-\frac{\sqrt{x}+15}{\sqrt{x}-3}+\frac{3\sqrt{x}-1}{\sqrt{x}-5}\)
\(=\frac{8\sqrt{x}-x-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}-\)\(\frac{\left(\sqrt{x}+15\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)\(+\frac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{8\sqrt{x}-x-31-x-10\sqrt{x}+75+3x-10\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{x-12\sqrt{x}+47}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(\Rightarrow\)Sai đề không cậu ưi
cho biểu thức M=\(\left(\frac{x+2\sqrt{x}+4}{x\sqrt{x}-8}+\frac{x+2\sqrt{x}+1}{x-1}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-2}+\frac{2\sqrt{x}+10}{x+6\sqrt{x}+5}\right)\)
Rút gọn
Rút gọn biểu thức \(\sqrt {x\sqrt {x\sqrt x } } :{x^{\frac{5}{8}}}(x > 0)\) ta được
A. \(\sqrt[4]{x}\)
B. \(\sqrt x \).
C. \(\sqrt[3]{x}\).
D. \(\sqrt[5]{x}\)
\(=\sqrt{x\sqrt{x^{1+\dfrac{1}{2}}}}:x^{\dfrac{5}{8}}\)
\(=\sqrt{x\cdot x^{\dfrac{1}{2}\cdot\dfrac{3}{2}}}:x^{\dfrac{5}{8}}\)
\(=\sqrt{x^{1+\dfrac{3}{4}}}:x^{\dfrac{5}{8}}\)
\(=x^{\dfrac{1}{2}\cdot\dfrac{7}{4}}:x^{\dfrac{5}{8}}=x^{\dfrac{7}{8}-\dfrac{5}{8}}=x^{\dfrac{1}{4}}=\sqrt[4]{x}\)
=>A
Rút gọn A=\(\left(\frac{5\sqrt{x}+50}{x+5\sqrt{x}}+\frac{2\sqrt{x}-10}{\sqrt{x}}+\frac{x}{5\sqrt{x}+25}\right).\frac{7}{15+3\sqrt{x}}\)
Bài 1: Rút gọn
a. \(\left(5-2\sqrt{3}\right)^2+\left(5+2\sqrt{3}\right)^2\)
b. \(\left(\sqrt{5}+\sqrt{2}\right)^2-\left(2\sqrt{5}+1\right)\left(2\sqrt{5}-1\right)-\sqrt{40}\)
c. \(\left(\sqrt{2}-1\right)^2-\frac{2}{3}\sqrt{4}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{15}}-\sqrt{2}\)
d. \(\left(\sqrt{6}-\sqrt{18}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}+2\sqrt{3}\)
e. \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+6\sqrt{6}+3\sqrt{24}\)
Bài 2: Rút gọn
A =\(\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}:\frac{\sqrt{x+1}}{x-2\sqrt{x}+1}\right)\)(x>0 ; x khác 1)
rút gọn Q= ($\frac{\sqrt{x+2} }{x-2\sqrt{x}+4 }$ - $\frac{x-\sqrt{x} }{x\sqrt{x} +8 }$ ). $\frac{5x-10\sqrt{x}+20 }{5\sqrt{x}+4}$
Tử số của phân số đầu phải là \(\sqrt{x}+2\) chứ không phải \(\sqrt{x+2}\), vì cái \(\sqrt{x}+2\) nó mới logic để rút gọn: )
\(Q=\left(\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}^3+8}-\dfrac{x-\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\left(\dfrac{x+4\sqrt{x}+4-x+\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\dfrac{\left(5\sqrt{x}+4\right).5.\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)\left(5\sqrt{x}+4\right)}\\ =\dfrac{5}{\sqrt{x}+2}\)
a,Cho biểu thức:\(M=\left(\frac{x+2\sqrt{x}+4}{x\sqrt{x}-8}+\frac{x+2\sqrt{x}+1}{x-1}\right):\left(\frac{3\sqrt{x}-5}{\sqrt{x}-2}+\frac{2\sqrt{x}+10}{x+6\sqrt{x}5}\right)\)
Rút gọn M và tìm x để M>1