7 + 2 x = 45 : 3
tìm x
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\(\dfrac{9}{7}x+\dfrac{5}{7}x=\dfrac{2}{3}\)
\(\Leftrightarrow x\left(\dfrac{9}{7}+\dfrac{5}{7}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{14}{7}x=\dfrac{2}{3}\)
\(\Leftrightarrow2x=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{2}{3}:2\)
\(\Leftrightarrow x=\dfrac{2}{3}\times\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{2}{6}\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Để hệ phương trình có nghiệm duy nhất thì \(\dfrac{m}{2}\ne\dfrac{2}{-4}=-\dfrac{1}{2}\)
=>\(m\ne-1\)
\(\left\{{}\begin{matrix}mx+2y=1\\2x-4y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2mx+4y=2\\2x-4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(2m+2\right)=5\\2x-4y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5}{2m+2}\\4y=2x-3=\dfrac{10}{2m+2}-3=\dfrac{10-6m-6}{2m+2}=\dfrac{-6m+4}{2m+2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5}{2m+2}\\y=\dfrac{-6m+4}{8m+8}=\dfrac{-3m+2}{4m+4}\end{matrix}\right.\)
x-3y=7/2
=>\(\dfrac{5}{2m+2}-\dfrac{3\cdot\left(-3m+2\right)}{4m+4}=\dfrac{7}{2}\)
=>\(\dfrac{10+3\left(3m-2\right)}{4m+4}=\dfrac{7}{2}\)
=>\(\dfrac{10+9m-6}{4m+4}=\dfrac{7}{2}\)
=>\(\dfrac{9m+4}{4m+4}=\dfrac{7}{2}\)
=>7(4m+4)=2(9m+4)
=>28m+28=18m+8
=>10m=-20
=>m=-2(nhận)
Ta có : \(\dfrac{x}{2}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x=2\left(1-x\right)\)
\(\Leftrightarrow3x=2-2x\)
\(\Leftrightarrow5x=2\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy ...
Ta có: \(\dfrac{x}{2}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x=2\left(1-x\right)\)
\(\Leftrightarrow3x=2-2x\)
\(\Leftrightarrow3x+2x=2\)
\(\Leftrightarrow5x=2\)
hay \(x=\dfrac{2}{5}\)
Vậy: \(x=\dfrac{2}{5}\)
\(\Leftrightarrow2.5:x=2+\dfrac{2}{3}-2-\dfrac{1}{5}=\dfrac{7}{15}\)
hay \(x=\dfrac{5}{2}:\dfrac{7}{15}=\dfrac{5}{2}\cdot\dfrac{15}{7}=\dfrac{75}{14}\)
a: A=x^2-2x+1+4
=(x-1)^2+4>=4
Dấu = xảy ra khi x=1
b: =x^2-x+1/4+3/4
=(x-1/2)^2+3/4>=3/4
Dấu = xảy ra khi x=1/2
c: =2x+8-x^2-4x
=-x^2-2x+8
=-x^2-2x-1+9
=-(x^2+2x+1)+9
=-(x+1)^2+9<=9
Dấu = xảy ra khi x=-1
d: =x^2-2xy+y^2+4y^2+4y+1+2
=(x-y)^2+(2y+1)^2+2>=2
Dấu = xảy ra khi x=y và 2y+1=0
=>x=y=-1/2
\(\)đặt \(2x^2+y^2+\dfrac{28}{x}+\dfrac{1}{y}=A\)
\(=>A=2x^2+y^2-7x-y+\dfrac{28}{x}+7x+\dfrac{1}{y}+y\)
\(A=2x^2-8x+8+y^2-2y+1+x+y-9+\dfrac{28}{x}+7x+\dfrac{1}{y}+y\)
\(A=2\left(x-2\right)^2+\left(y-1\right)^2+\left(x+y\right)-9+\dfrac{28}{x}+7x+\dfrac{1}{y}+y\)
áp dụng BDT AM-GM\(=>\dfrac{28}{x}+7x+\dfrac{1}{y}+y\ge2\sqrt{28.7}+2\sqrt{1}=30\)
\(=>A\ge30+3-9=24\)
dấu"=" xảy ra<=>x=2,y=1
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\\x=-1\end{matrix}\right.\)
7 + 2x = 45 : 3
7 + 2x = 15
2x = 15 - 7
2x = 8
x = 8 : 2
x = 4
Vậy x = 4
TA CÓ : 7+2x=45:3<=>7+2x=15<=>2x=15-7<=>2x=8<=>x=4
vậy x=4