Bài 1 Tính giá trị biểu thức
a. \(\sqrt{0,16}\)+ \(\sqrt{0,04}\) - \(\sqrt{0,25}\)
b.\(\sqrt{85^2-84^2}\) - \(\sqrt{26^2-24^2}\)
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RK
12 tháng 7 2021
a) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\\\Rightarrow2\sqrt{31}>10\)
15 tháng 6 2023
\(=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{x+16}{\sqrt{x}+3}\)
Khi x=căn 2 thì \(A=\dfrac{\sqrt{2}+16}{\sqrt{\sqrt{2}}+3}\)
8 tháng 2 2023
a: \(A=5\sqrt{2}-6\sqrt{2}+\sqrt{2}-1=-1\)
\(B=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}=\dfrac{x+\sqrt{x}}{x-1}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: A=B
=>căn x=-căn x+1
=>căn x=1/2
=>x=1/4
15 tháng 12 2023
a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=5-1=4\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=-\dfrac{2}{\sqrt{x}+1}\)
c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)
d: |B|=A
=>\(\left|-\dfrac{2}{\sqrt{x}+1}\right|=4\)
=>\(\dfrac{2}{\sqrt{x}+1}=4\) hoặc \(\dfrac{2}{\sqrt{x}+1}=-4\)
=>\(\sqrt{x}+1=\dfrac{1}{2}\) hoặc \(\sqrt{x}+1=-\dfrac{1}{2}\)
=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)
18 tháng 8 2023
a: Khi x=25 thì \(A=\dfrac{7}{5+8}=\dfrac{7}{13}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{x-9}\)
\(=\dfrac{x+5\sqrt{x}-24}{x-9}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{x-9}=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)
c: P=A*B
\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\cdot\dfrac{7}{\sqrt{x}+8}=\dfrac{7}{\sqrt{x}+3}\)
P là số nguyên
=>căn x+3 thuộc Ư(7)
=>căn x+3=7
=>x=16
1 tháng 9 2021
Bài 2:
Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
2
14 tháng 5 2023
`a)A=[2\sqrt{3}+2-2\sqrt{3}+2]/[(2\sqrt{3}-2)(2\sqrt{3}+2)]`
`A=4/[12-4]=1/2`
Với `x > 0,x ne 1` có:
`B=[x-2\sqrt{x}+1]/[\sqrt{x}(\sqrt{x}-1)]`
`B=[(\sqrt{x}-1)^2]/[\sqrt{x}(\sqrt{x}-1)]=[\sqrt{x}-1]/\sqrt{x}`
`b)B=2/5A`
`=>[\sqrt{x}-1]/\sqrt{x}=2/5 . 1/2`
`<=>5\sqrt{x}-5=\sqrt{x}`
`<=>\sqrt{x}=5/4`
`<=>x=25/16` (t/m)
23 tháng 10 2021
a: \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\sqrt{x}-1\)
23 tháng 10 2021
a) \(A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
Đk: \(x>0\) và \(x\ne1\)
\(\Rightarrow A=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{x-\sqrt{x}}\)
\(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x\sqrt{x}-2x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(x-2\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
b) Thay \(x=3+2\sqrt{2}\) vào A ta được:
\(A=\sqrt{3+2\sqrt{2}}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1\)
\(=\sqrt{2}+1-1=\sqrt{2}\)
(Vì \(\sqrt{2}+1>0\Rightarrow\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\))
H
27 tháng 10 2023
Bài `1`
\(\sqrt{4-2\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{\sqrt{3}-3}{\sqrt{3}-1}\\ =\sqrt{3-2\sqrt{3}+1}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}-\dfrac{2\left(\sqrt{3}-1\right)}{2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}+1-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}+1-\sqrt{3}\\ =-\sqrt{3}\)
27 tháng 10 2023
2:
a: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)
b: B=5
=>\(5\left(\sqrt{x}+3\right)=\sqrt{x}+8\)
=>\(5\sqrt{x}+15=\sqrt{x}+8\)
=>\(4\sqrt{x}=-7\)(loại)
Vậy: \(x\in\varnothing\)
a) \(\sqrt{0,16}+\sqrt{0,04}-\sqrt{0,25}\)
= 0,4 + 0,2 - 0,5
= 0,1
b) \(\sqrt{85^2-84^2}-\sqrt{26^2-24^2}\)
= \(\sqrt{\left(85-84\right)\left(85+84\right)}\) - \(\sqrt{\left(26-24\right)\left(26+24\right)}\)
= \(\sqrt{169}\) - \(\sqrt{2.50}\)
= 13 - 10
= 3
Chúc bạn học tốt
a) Ta có: \(\sqrt{0.16}+\sqrt{0.04}-\sqrt{0.25}\)
\(=0,4+0,2-0,5\)
=0,1