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Hệ có nghiệm duy nhất \(\Leftrightarrow\left(a+1\right)\left(a-1\right)\ne-1\Leftrightarrow a^2\ne0\) hay a ≠​ 0

Hệ \(\Leftrightarrow\left\{{}\begin{matrix}a\left(x-y\right)=a-3\\x+\left(a-1\right)y=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\left(a-3\right)}{a}+y\\\left(\frac{\left(a-3\right)}{a}+y\right)+\left(a-1\right)y=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\left(a-3\right)}{a}+y\\\frac{\left(a-3\right)}{a}+ay=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\left(a-3\right)}{a}+\frac{a+3}{a^2}\\y=\frac{a+3}{a^2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{a^2-2a+3}{a^2}\\y=\frac{a+3}{a^2}\end{matrix}\right.\)

=> x+y=\(\frac{a^2-a+6}{a^2}=1-\frac{1}{a}+6.\frac{1}{a^2}\)
Đặt \(\frac{1}{a}=t\)
=> 6t²-t+1=\(6\left(t-\frac{1}{12}\right)^2+\frac{23}{24}\ge\frac{23}{24}\)
Dấu bằng xảy ra khi và chỉ khi \(t-\frac{1}{12}=0\Leftrightarrow t=\frac{1}{12}\Leftrightarrow a=12\)

\(x-5=0\)

\(x=0+5=5\)

\(x\cdot7=14\)

\(x=\frac{14}{7}=2\)

\(x+2=6\)

\(x=6-2=4\)

số em trong lớp đó là :

10+10+4=24 em

       Đáp số 24 em

126cm2

a. \(x-5=0\)

\(x=0+5\)

\(x=5\)

b. \(x.7=14\)

\(x=14:7\)

\(x=2\)

c.  \(x+2=6\)

\(x=6-2\)

\(x=4\)

a) x - 5 = 0

=> x     = 0 + 5

=> x      = 5

b) x . 7 = 14 

=> x     = 14 : 7

=> x      = 2

c) x + 2 = 6

=> x      = 6-2

=> x       = 4

Ai k mik mik k lại

=3kg rau