1) Cho \(\left(x+\sqrt{x^2+2}\right)\left(y+\sqrt{y^2+2}\right)=2\)
Tinh S=\(x\sqrt[]{y^2+2}+y\sqrt{y^2+2}\)
2) Cho a+b+c=0 & \(a^2+b^2+c^2=1\)
Tinh M=\(a^4+b^4+c^4\)
Những câu hỏi liên quan
Cho x,y>0 tm xy+x+y=1. Tính
\(S=x\sqrt{\frac{2\left(1+y^2\right)}{1+x^2}}+y\sqrt{\frac{2\left(1+x^2\right)}{1+y^2}}+\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{2}}\)
Cho 3 số dương x,y,z thỏa mãn x + y + z = xyz. Cmr:
\(A=\frac{\sqrt{\left(1+y^2\right)\left(1+z^2\right)}-\sqrt{1+y^2}-\sqrt{1+z^2}}{yz}+\frac{\sqrt{\left(1+z^2\right)\left(1+x^2\right)}-\sqrt{1+x^2}-\sqrt{1+z^2}}{xz}+\frac{\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\sqrt{1+x^2}-\sqrt{1+y^2}}{xy}=0\)
@Akai Haruma, Nguyen, Nguyễn Thị Ngọc Thơsvtkvtm
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Câu hỏi của Vũ Sơn Tùng - Toán lớp 9 | Học trực tuyến
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1) cho a,b,c dương thỏa a+b+c1 CMR sqrt{left(ab+cright)left(bc+aright)left(ac+bright)}left(1-aright)left(1-bright)left(1-cright)
2) cho x,y dương thỏa mãn xsqrt{x}+ysqrt{y}x^2+y^2x^2sqrt{x}+y^2sqrt{y} .tính tổng x+y
3) ghpt left{{}begin{matrix}x^2+2y^223x^2+4xy+4x+3yy^2-4end{matrix}right.
4) gpt sqrt{x^2+3}+dfrac{4x}{sqrt{x^2+3}}5sqrt{x}
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1) cho a,b,c dương thỏa a+b+c=1 CMR \(\sqrt{\left(ab+c\right)\left(bc+a\right)\left(ac+b\right)}=\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
2) cho x,y dương thỏa mãn \(x\sqrt{x}+y\sqrt{y}=x^2+y^2=x^2\sqrt{x}+y^2\sqrt{y}\) .tính tổng x+y
3) ghpt \(\left\{{}\begin{matrix}x^2+2y^2=2\\3x^2+4xy+4x+3y=y^2-4\end{matrix}\right.\)
4) gpt \(\sqrt{x^2+3}+\dfrac{4x}{\sqrt{x^2+3}}=5\sqrt{x}\)
gợi ý nè
1) \(ab+c=ab+c\left(a+b+c\right)\)....
2) nhiều cách lắm nhưng tớ chỉ đưa ra 2 cách ...có vẻ hay
đặt \(\sqrt{x}=a,\sqrt{y}=b\)
=>a3+b3=a4+b4=a5+b5
c1: ta có: \(\left(a^3+b^3\right)\left(a^5+b^5\right)=\left(a^4+b^4\right)^2\)......
c2: a5+b5=(a+b)(a4+b4)-ab(a3+b3)
=> 1=(a+b)-ab .......
3) try use UCT
4) tính sau =))
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Cho x,y >0. Rút gọn A=\(\sqrt{2\left(\sqrt{x^2+y^2}+x\right)\left(\sqrt{x^2+y^2}+y\right)}-\sqrt{x^2+y^2}-x-y+2020\)
\(A=\sqrt{2\left(x^2+y^2+\left(x+y\right)\sqrt{x^2+y^2}+xy\right)}-\sqrt{x^2+y^2}-x-y+2020\)
\(=\sqrt{\left(x^2+y^2+2xy\right)+x^2+y^2+2\left(x+y\right)\sqrt{x^2+y^2}}-\sqrt{x^2+y^2}-x-y+2020\)
\(=\sqrt{\left(x+y\right)^2+2\left(x+y\right)\sqrt{x^2+y^2}+x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020\)
\(=\sqrt{\left(x+y+\sqrt{x^2+y^2}\right)^2}-\sqrt{x^2+y^2}-x-y+2020\)
\(=x+y+\sqrt{x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020\)
\(=2020\)
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a:dfrac{b}{left(a-4right)^2}.sqrt{dfrac{left(a-4right)^4}{b^2}}left(b0;ane4right)
b:dfrac{xsqrt{x}-ysqrt{y}}{sqrt{x}-sqrt{y}}left(xge0;yge0;xne0right)
c:dfrac{a}{left(b-2right)^2}.sqrt{dfrac{left(b-2right)^4}{a^2}left(a0;bne2right)}
d:dfrac{x}{left(y-3right)^2}.sqrt{dfrac{left(y-3right)^2}{x^2}left(x0;yne3right)}
e:2x +dfrac{sqrt{1-6x+9x^2}}{3x-1}
Đọc tiếp
a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
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a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
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Cho x;y;z>0 xy+yz+xz=1 tinh
A= \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
cho x;y;z>0 xy+yz+xz=1 tinh\(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
ta có đăng thức a2 +ab+bc+ca=(a+b)(a+c)
theo đề suy ra a^2 +1=(a+b)(a+c)
khúc này bạn tự làm típ
suy ra biểu thức trên bằng a(b+c)+b(a+c)+c(a+b)
=2(ab+ac+bc)=2
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Cho 2 số thực a, b thỏa mãn xy + \(\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=1\)
CMR: \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=0\)
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!!