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\(\sqrt{5+3x}-\sqrt{5-3x}=a\left(x\le\dfrac{5}{3}\right)\)
\(\Rightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\)
\(\Rightarrow5+3x+5-3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}=a^2\)
\(\Rightarrow10-2\sqrt{25-9x^2}=a^2\)
\(\Rightarrow-2\sqrt{25-9x^2}=a^2-10\)
\(\Rightarrow2\sqrt{25-9x^2}=10-a^2\)
\(\Rightarrow10+2\sqrt{25-9x^2}=20-a^2\)
\(\Rightarrow P=\dfrac{\sqrt{10+2\sqrt{25-9x^2}}}{x}=\dfrac{\sqrt{20-a^2}}{x}\)
\(\sqrt{5+3x}-\sqrt{5-3x}=a\\ \Rightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\\ \Rightarrow5+3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}+5-3x=a^2\\ \Rightarrow2\sqrt{25-9x^2}=10-a^2\\ \Rightarrow4\left(25-9x^2\right)=\left(10-a^2\right)^2\\ \Rightarrow100-36x^2=100-20a^2+a^4\\ \Rightarrow36x^2=20a^2-a^4\\ \Rightarrow x^2=\dfrac{20a^2-a^4}{36}\\ \Rightarrow x=\dfrac{\sqrt{a^2\left(20-a^2\right)}}{6}\)
\(\Rightarrow P=\dfrac{\sqrt{10+2\sqrt{25-9x^2}}}{x}\\ =\dfrac{\sqrt{10+10-a^2}}{\dfrac{\sqrt{a^2\left(20-a^2\right)}}{6}}=6\sqrt{\dfrac{20-a^2}{a^2\left(20-a^2\right)}}=\dfrac{6}{\left|a\right|}\)
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
1) Sửa đề: x=0,09
Thay x=0,09 vào A, ta được:
\(A=\dfrac{\sqrt{0.09}}{\sqrt{0.09}-1}=\dfrac{0.3}{0.3-1}=\dfrac{0.3}{-0.7}=\dfrac{-3}{7}\)
Bài 1:
Thay x=9 vào biểu thức \(A=\frac{2\sqrt{x}+1}{\sqrt{x}+2}\), ta được:
\(\frac{2\cdot\sqrt{9}+1}{\sqrt{9}+2}=\frac{2\cdot3+1}{3+2}=\frac{7}{5}\)
Vậy: \(\frac{7}{5}\) là giá trị của biểu thức \(A=\frac{2\sqrt{x}+1}{\sqrt{x}+2}\) tại x=9
Bài 2:
a) Ta có: \(B=\left(\frac{x+14\sqrt{x}-5}{x-25}+\frac{\sqrt{x}}{\sqrt{x}+5}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)
\(=\left(\frac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right)\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(=\frac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(=\frac{2x+10\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\cdot\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}-1}{\sqrt{x}+2}\)
a) Ta có: \(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
P=\(\frac{\sqrt{10+2\sqrt{25-9x^2}}}{x}\)
P=\(\frac{\sqrt{10+2\sqrt{\left(5+3x\right)\left(5-3x\right)}}}{x}\)
P=\(\frac{\sqrt{10+10-a^2}}{x}\)(Vì a2=\(\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2\)=10-2\(\sqrt{\left(5+3x\right)\left(5-3x\right)}\))
\(\sqrt{5+3x}-\sqrt{5-3x}=a\)
\(\Leftrightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\)
\(\Leftrightarrow5+3x+5-3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}=a^2\)
\(\Leftrightarrow10-2\sqrt{\left(5+3x\right)\left(5-3x\right)}=a^2\)
\(\Leftrightarrow2\sqrt{\left(5+3x\right)\left(5-3x\right)}=10-a^2\)
Thế vào P ta được:
\(P=\frac{\sqrt{10+2\sqrt{25-9x^2}}}{x}=\frac{\sqrt{10+2\sqrt{\left(5-3x\right)\left(5+3x\right)}}}{x}\)
\(=\frac{\sqrt{10+10-a^2}}{x}\)
\(=\frac{\sqrt{20-a^2}}{x}\)
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