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a: A=x+3+|x-3|
=x+3+3-x(x<=3)
=6
b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\left|x+2\right|-\left|x\right|\)
=x+2-x=2
c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)
\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)
Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ bạn tốt hơn nhé.
a: Khi x=64 thì \(A=\dfrac{2}{8-2}=\dfrac{2}{6}=\dfrac{1}{3}\)
b: \(P=B:A\)
\(=\dfrac{3\sqrt{x}+\sqrt{x}-2-2\left(\sqrt{x}+2\right)}{x-4}:\dfrac{2}{\sqrt{x}-2}\)
\(=\dfrac{4\sqrt{x}-2-2\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}\)
\(=\dfrac{2\sqrt{x}-6}{2\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
c: P<0
=>căn x-3<0
=>0<=x<9
mà x nguyên và x<>4
nên \(x\in\left\{0;1;2;3;5;6;7;8\right\}\)
`a)|x-2|=2<=>[(x=4(ko t//m)),(x=0(t//m)):}`
Thay `x=0` vào `A` có: `A=[2\sqrt{0}-3]/[\sqrt{0}-2]=3/2`
`b)` Với `x >= 0,x ne 4` có:
`B=[2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[2\sqrt{x}-6+x+3\sqrt{x}-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[x+\sqrt{x}-6]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[(\sqrt{x}+3)(\sqrt{x}-2)]/[(\sqrt{x}+3)(\sqrt{x}-3)]`
`B=[\sqrt{x}-2]/[\sqrt{x}-3]`
`c)` Với `x >= 0,x ne 4` có:
`C=A.B=[2\sqrt{x}-3]/[\sqrt{x}-2].[\sqrt{x}-2]/[\sqrt{x}-3]=[2\sqrt{x}-3]/[\sqrt{x}-3]`
Có: `C >= 1`
`<=>[2\sqrt{x}-3]/[\sqrt{x}-3] >= 1`
`<=>[2\sqrt{x}-3-\sqrt{x}+3]/[\sqrt{x}-3] >= 0`
`<=>[\sqrt{x}]/[\sqrt{x}-3] >= 0`
Vì `x >= 0=>\sqrt{x} >= 0`
`=>\sqrt{x}-3 > 0`
`<=>x > 9` (t/m đk)
a: Thay x=2 vào B, ta được:
\(B=\dfrac{2}{\sqrt{2}-1}=2\sqrt{2}+2\)
a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)
\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)
\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)
\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)
c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)
\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)
\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)
\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)
d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
a: Sửa đề: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
Khi x=9 thì \(B=\dfrac{\sqrt{9}+1}{\sqrt{9}+2}\)
\(=\dfrac{3+1}{3+2}=\dfrac{4}{5}\)
b: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{6+\sqrt{x}}{x-4}\)
\(=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+6}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-5\sqrt{x}+6+x+2\sqrt{x}-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x-4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}+2}\)
c: P=A/B
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
\(P-2=\dfrac{2\sqrt{x}}{\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{-2}{\sqrt{x}+1}< 0\)
=>P<2
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
