a.\(3\left(x-1\right)=3\left(y-2\right);4\left(y-2\right)=3\left(z-3\right)v\text{à}2x+3y-z=-250\)
b.\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}v\text{à}x^2+y^2+z^2=14\)
giải ra giúp mik nha!!!!!!!!!
Những câu hỏi liên quan
giải hệ phương trình a)\(\left\{{}\begin{matrix}2\left(x+1\right)-3\left(y-2\right)=5\\-4\left(x-2\right)+5\left(y-3\right)=-1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}8\left(x-3\right)-3\left(y+1\right)=-2\\3\left(x+2\right)-2\left(1-y\right)=5\end{matrix}\right.\)
Help me ~~~
a) Ta có: \(\left\{{}\begin{matrix}2\left(x+1\right)-3\left(y-2\right)=5\\-4\left(x-2\right)+5\left(y-3\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2-3y+6=5\\-4x+8+5y-15=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\2x-3y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x-3\cdot0=-3\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)
Vậy: hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}8\left(x-3\right)-3\left(y+1\right)=-2\\3\left(x+2\right)-2\left(1-y\right)=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x-24-3y-3=-2\\3x+6-2+2y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y=75\\24x+16y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-25y=67\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-67}{25}\\3x=1-2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=1-2\cdot\dfrac{-67}{25}=\dfrac{159}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
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a) HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\x=\dfrac{3y-3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(-\dfrac{3}{2};0\right)\)
b) HPT \(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}16x-6y=50\\9x+6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}25x=53\\y=\dfrac{1-3x}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\)
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Rút gọn biểu thức:a) Aleft(x-yright)^3+left(y+xright)^3+left(y-xright)^3-3xyleft(x+yright)b) B3x^2left(x+1right)left(x-1right)-left(x^2-1right)left(x^4+x^2+1right)+left(x^2-1right)^3c) Cleft(x+yright)left(x^2-xy+y^2right)+left(x-yright)left(x^2+xy+y^2right)-2x^3d) Dleft(x+1right)^3+left(x-1right)^3+x^3-3xleft(x+1right)left(x-1right)
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Rút gọn biểu thức:
a) \(A=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
b) \(B=3x^2\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x^4+x^2+1\right)+\left(x^2-1\right)^3\)
c) \(C=\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x^2+xy+y^2\right)-2x^3\)
d) \(D=\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)
Rút gọn biểu thức:
a) Aleft(x-yright)^3+left(y+xright)^3+left(y-xright)^3-3xyleft(x+yright)
b) B3x^2left(x+1right)left(x-1right)-left(x^2-1right)left(x^4+x^2+1right)+left(x^2-1right)^3
c) Cleft(x+yright)left(x^2-xy+y^2right)+left(x-yright)left(x^2+xy+y^2right)-2x^3
d) Dleft(x+1right)^3+left(x-1right)^3+x^3-3xleft(x+1right)left(x-1right)
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Rút gọn biểu thức:
a) \(A=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
b) \(B=3x^2\left(x+1\right)\left(x-1\right)-\left(x^2-1\right)\left(x^4+x^2+1\right)+\left(x^2-1\right)^3\)
c) \(C=\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x^2+xy+y^2\right)-2x^3\)
d) \(D=\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)
thực hiện phép tính
a,x^3+left[frac{xleft(2y^3-x^3right)}{x^3+y^3}right]^3-left[frac{yleft(2x^3-y^3right)}{x^3+y^3}right]^3
b,frac{frac{xleft(x+yright)}{x-y}+frac{xleft(x+zright)}{x-z}}{1+frac{left(y-zright)^2}{left(x-yright)left(x-zright)}}+frac{frac{yleft(y+zright)}{y-z}+frac{yleft(y+xright)}{y-x}}{1+frac{left(z-xright)^2}{left(y-zright)left(y-xright)}}+frac{frac{zleft(z+xright)}{z-x}+frac{zleft(z+yright)}{z-y}}{1+frac{left(x-yright)^2}{left(z-xright)left(z-yright)}}
c,left[frac{y+z-2x}{fra...
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thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
Cho x, y, z là các số thực bất kì. Chứng minh rằng:
a) \(\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)\ge\left(xy+yz+zx-1\right)^2\)
b) \(\left(x^2+2\right)\left(y^2+2\right)\left(z^2+2\right)\ge3\left(x+y+z\right)^2\)
c) \(\left(x^3+3\right)\left(y^3+3\right)\left(z^3+3\right)\ge4\left(x+y+z+1\right)^2\)
1)tìm các số nguyên x và y thỏa mãn:y^2x^2+x+12)cho các số thực x và y thỏa mãn left(x+sqrt{a+x^2}right)left(y+sqrt{a+y^2}right)atìm giá trị biểu thức 4left(x^7+y^7right)+2left(x^5+y^5right)+11left(x^3+y^3right)+20163)cho x;y là các số thực khác 0 thỏa mãn x+y khác 0cmr frac{1}{left(x+yright)^3}left(frac{1}{x^3}+frac{1}{y^3}right)+frac{3}{left(x+yright)^4}left(frac{1}{x^2}+frac{1}{y^2}right)+frac{6}{left(x+yright)^5}left(frac{1}{x}+frac{1}{y}right)frac{1}{x^3y^3}4)cho a,b,c là các số dương.cmrsq...
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1)tìm các số nguyên x và y thỏa mãn:\(y^2=x^2+x+1\)
2)cho các số thực x và y thỏa mãn \(\left(x+\sqrt{a+x^2}\right)\left(y+\sqrt{a+y^2}\right)\)=a
tìm giá trị biểu thức \(4\left(x^7+y^7\right)+2\left(x^5+y^5\right)+11\left(x^3+y^3\right)+2016\)
3)cho x;y là các số thực khác 0 thỏa mãn x+y khác 0
cmr \(\frac{1}{\left(x+y\right)^3}\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\frac{3}{\left(x+y\right)^4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\frac{6}{\left(x+y\right)^5}\left(\frac{1}{x}+\frac{1}{y}\right)\)\(=\frac{1}{x^3y^3}\)
4)cho a,b,c là các số dương.cmr\(\sqrt{\frac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\frac{b^3}{b^3+\left(a+c\right)^3}}+\sqrt{\frac{c^3}{c^3+\left(a+b\right)^3}}\ge1\)
\(a.\left|x+2\right|+\left|x-1\right|=3-\left(y+2\right)^2\)
\(b.\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
\(c.\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
\(d.\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y-3\right|+3}\)
a) Áp dụng bất đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có :
\(\left|x+2\right|+\left|x-1\right|=\left|x+2\right|+\left|1-x\right|\)
\(\ge\left|x+2+1-x\right|=3\) (1)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+2\right)\left(1-x\right)\ge0\)
\(\Leftrightarrow-2\le x\le1\)
+ \(\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow3-\left(y+2\right)^2\le3\) (2)
Dấu "=" xảy ra \(\Leftrightarrow\left(y+2\right)^2=0\Leftrightarrow y=-2\)
Từ (1) và (2) suy ra \(\left|x+2\right|+\left|x+1\right|=3-\left(y+2\right)^2=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le1\\y=-2\end{matrix}\right.\)
b) \(\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\) (3)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-5\right)\left(1-x\right)\ge0\)
\(\Leftrightarrow1\le x\le5\)
+ \(\left|y+1\right|\ge0\forall y\) \(\Rightarrow\left|y+1\right|+3\ge3\)
\(\Rightarrow\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\) (4)
Dấu "=" xảy ra \(\Leftrightarrow\left|y+1\right|=0\Leftrightarrow y=-1\)
Từ (3) và (4) suy ra \(\left|x-5\right|+\left|1-x\right|=\frac{12}{\left|y+1\right|+3}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-1\end{matrix}\right.\)
Câu c,d lm tương tự
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1.a, Giải phương trình \(\left(\sqrt{x 3}-\sqrt{x 1}\right)\left(x^2 \sqrt{x^2 4x 3}\right)=2x\)b, Giải hệ phương trình \(\left\{{}\begin{matrix}x^2\left(y^2 1\right) 2y\left(x^2 x 1\right)=3\\\left(x^2 x\right)\left(y^2 y\right)=1\end{matrix}\right....
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Rút gọn biểu thức
a)\(\left(x+y\right)^3+\left(x-y\right)^3-2x^3\)
b) \(\left(x+y\right)^2-\left(x-y\right)^2+\left(x+y\right)\left(x-y\right)\)
c)\(\left(3x+1\right)^2+2\left(9x^2-1\right)+\left(3x-1\right)^2\)
d) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)
a ) \(\left(x+y\right)^3+\left(x-y\right)^3-2x^3\)
\(=x^3+3x^2y+3y^2x+y^3+x^3-3x^2y+3y^2x-y^3-2x^3\)
\(=\left(x^3+x^3-2x^3\right)+\left(y^3-y^3\right)+\left(3x^2y-3x^2y\right)+\left(3y^2x+3y^2x\right)\)
\(=6y^2x\)
b ) \(\left(x+y\right)^2-\left(x-y\right)^2+\left(x+y\right)\left(x-y\right)\)
\(=\left(x+y-x+y\right)\left(x+y+x-y\right)+x^2-y^2\)
\(=2y.2x+x^2-y^2\)
\(=x^2-y^2+4xy\)
c ) \(\left(3x+1\right)^2+2\left(9x^2-1\right)+\left(3x-1\right)^2\)
\(=\left(3x+1\right)^2+2\left(3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(3x+1+3x-1\right)^2\)
\(=\left(6x\right)^2=36x^2\)
d ) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(a+b+c-b-c\right)^2\)
\(=a^2\)
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Rút gọn
a) \(x.\left(x+4\right).\left(x-4\right)-\left(x^2+1\right).\left(x-1\right)\)
b) \(\left(y-3\right).\left(y+3\right).\left(y^2+9\right)-\left(y^2+2\right).\left(y^2-2\right)\)
a) \(x\left(x^2-16\right)-\left(x^2+1\right)\left(x-1\right)\) =\(x^3-16x^2-x^3+x^2-x+1\)
= \(x^2-17x+1\)
b) \(\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\) = \(\left(y^4-81\right)-\left(y^4-16\right)\)
=\(-65\)
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