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!!!!!!!!!
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24 tháng 1 2021
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 \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
24 tháng 1 2021
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 \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\)
KB
18 tháng 9 2018
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 ) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(a+b+c-b-c\right)^2\)
\(=a^2\)
27 tháng 6 2016
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) \(\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\) = \(\left(y^4-81\right)-\left(y^4-16\right)\)
=\(-65\)
b. \(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Rightarrow\left(\frac{x}{2}\right)^3=\left(\frac{y}{4}\right)^3=\left(\frac{z}{6}\right)^3\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)
\(\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\)
Theo t/c dảy tỉ số = nhau:
\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)
=> \(\frac{x^2}{4}=\frac{1}{4}\Rightarrow x^2=\frac{1}{4}.4=1=1^2=\left(-1\right)^2\Rightarrow x=\)+1
=> \(\frac{y^2}{16}=\frac{1}{4}\Rightarrow y^2=\frac{1}{4}.16=4=2^2=\left(-2\right)^2\Rightarrow y=\)+2
=> \(\frac{z^2}{36}=\frac{1}{4}\Rightarrow z^2=\frac{1}{4}.36=9=3^2=\left(-3\right)^2\Rightarrow z=\)+3
Vậy có 2 cặp (x;y;z) là: (1;2;3) và (-1;-2;-3).
a. Áp dụng t/c tỉ số = nhau làm tương tự.