Câu 3: Phân tích ra thừa số:
a. \(\sqrt{xy}-x\)
b. \(x+y-2\sqrt{xy}\)
c. \(x\sqrt{y}-y\sqrt{x}\)
d. \(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6\)
Những câu hỏi liên quan
19 tháng 11 2021 lúc 7:49
Câu nào đúng trong các câu sau (với x, y không âm) ?
A. \(x\sqrt{y}-\sqrt{xy}=xy\left(1-\sqrt{xy}\right)\)
B. \(x\sqrt{y}-\sqrt{xy}=\sqrt{xy}\left(\sqrt{x}-1\right)\)
C. \(x\sqrt{y}-\sqrt{xy}=\sqrt{y}\left(x-1\right)\)
D. \(x\sqrt{y}-\sqrt{xy}=x\sqrt{y}\left(1-\sqrt{xy}\right)\)
Tìm điều kiện xác định và phân tích các đa thức sau thành nhân tử:
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(C=\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(D=\sqrt{x^2+3x+2}+\sqrt{x+1}+2\sqrt{x+2}+2\)
\(A,ĐKXĐ:x;y\ge0\)
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)
\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)
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\(ĐKXĐ:x;y\ge0\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)
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\(ĐKXĐ:x;y\ge0\)
\(C=\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(=\left(\sqrt{x^3}+\sqrt{x^2y}\right)-\left(\sqrt{y^3}+\sqrt{xy^2}\right)\)
\(=\sqrt{x^2}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{y^2}\left(\sqrt{y}+\sqrt{x}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{x}-\sqrt{y}\right)\)
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Xem thêm câu trả lời
Rút gọn các biểu thức sau:a) Adfrac{xsqrt{y}+ysqrt{x}}{x+2sqrt{xy}+y}(x≥0 , y≥0 , xy≠0)b) Bdfrac{xsqrt{y}-ysqrt{x}}{x-2sqrt{xy}+y}(x≥0 , y≥0 , x≠y)c) Cdfrac{3sqrt{a}-2a-1}{4a-4sqrt{a}+1}(a≥0 , a≠dfrac{1}{4})d) Ddfrac{a+4sqrt{a}+4}{sqrt{a}+2}+dfrac{4-a}{sqrt{a}-2}(a≥0 , a≠4)
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Rút gọn các biểu thức sau:
a) A=\(\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)(x≥0 , y≥0 , xy≠0)
b) B=\(\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)(x≥0 , y≥0 , x≠y)
c) C=\(\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)(a≥0 , a≠\(\dfrac{1}{4}\))
d) D=\(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)(a≥0 , a≠4)
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) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)
\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)
\(D=0\)
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Phân tích đa thức thành nhân tử (với các căn thức đã cho đều có nghĩa)
A = \(x-y-3\left(\sqrt{x}+\sqrt{y}\right)\)
B = \(x-4\sqrt{x}+4\)
C = \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
D = \(5x^2-7x\sqrt{y}+2y\)
Giải hệ pt
1/left{{}begin{matrix}4xsqrt{y+1}+8xleft(4x^2-4x-3right)sqrt{x+1}dfrac{x}{x+1}+x^2left(y+2right)sqrt{left(x+1right)left(y+1right)}end{matrix}right.
2/left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.
3/left{{}begin{matrix}left(2x+y-1right)left(sqrt{x+3}+sqrt{xy}+sqrt{x}right)8sqrt{x}left(sqrt{x+3}+sqrt{xy}right)^2+xy2xleft(6-xright)end...
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Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!
Cm
a.\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\)
b. \(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{x}}{y-\sqrt{xy}}=-\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)với x>0, y>o và x≠y
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
\(=\dfrac{1}{6}\sqrt{6}\)
b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
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Phân tích thành nhân tử a) \(\sqrt{xy}-x\) b) \(x+y-2\sqrt{xy}\) c) \(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}\)-6 d) \(x\sqrt{y}-y\sqrt{x}\)
Câu a : \(\sqrt{xy}-x=\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)\)
Câu b : \(x+y-2\sqrt{xy}=\left(\sqrt{x}-\sqrt{y}\right)^2=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)
Câu c : \(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=\sqrt{x}\left(\sqrt{y}+2\right)-3\left(\sqrt{y}+2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{y}+2\right)\)
Câu d : \(x\sqrt{y}-y\sqrt{x}=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)
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1. Tính:
sqrt{dfrac{x-1+sqrt{2x-3}}{x+2-sqrt{2x+3}}}
2. Chứng minh:
a) dfrac{left(3sqrt{xy}-6y.2xsqrt{y}+4ysqrt{x}right)left(3sqrt{y}+2sqrt{xy}right)}{yleft(sqrt{x}-2sqrt{y}right)left(y-4xright)}1
b) left(sqrt{x}-sqrt{y}-dfrac{sqrt{xy}}{sqrt{x}+sqrt{y}}right)left(dfrac{sqrt{x}}{sqrt{x}+sqrt{y}}+dfrac{y}{sqrt{x}-sqrt{y}}-dfrac{2sqrt{xy}}{xy}right)sqrt{x}+sqrt{y}
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1. Tính:
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\)
2. Chứng minh:
a) \(\dfrac{\left(3\sqrt{xy}-6y.2x\sqrt{y}+4y\sqrt{x}\right)\left(3\sqrt{y}+2\sqrt{xy}\right)}{y\left(\sqrt{x}-2\sqrt{y}\right)\left(y-4x\right)}=1\)
b) \(\left(\sqrt{x}-\sqrt{y}-\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{y}{\sqrt{x}-\sqrt{y}}-\dfrac{2\sqrt{xy}}{xy}\right)=\sqrt{x}+\sqrt{y}\)
1.
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)
hết tối giải rồi
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CM:
\(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{6}}{6}\)
\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}+\sqrt{y}}=x-y\) với x.0, y>0, x≠y
\(\dfrac{\sqrt{y}}{x-\sqrt{xy}}+\dfrac{\sqrt{x}}{y-\sqrt{xy}}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)với x>0, y>0, x≠y
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
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