Tìm x trong các tỉ lệ thức
a, 2,5:7,5=x:3/5
b,x:5=3:7
c,x:37=-2/3,6
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a)2,5:7,5=x:3/5
0,(3)=x:3/5
\(\Rightarrow\)x= 1/5
b)x:2,5=0,003;0,75
x:2,5=1/250
\(\Rightarrow\)x=1/100
a) \(\left(\dfrac{1}{4}x\right):3=\dfrac{5}{6}:\dfrac{1}{8}\\ \left(\dfrac{1}{4}x\right):3=\dfrac{20}{3}\\ \dfrac{1}{4}x=\dfrac{20.3}{3}\\ \dfrac{1}{4}x=20\\ x=80\)
a: Ta có: \(\dfrac{0.25x}{3}=\dfrac{5.6}{0.125}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=134.4\)
hay x=537,6
b: Ta có: \(\dfrac{2.5}{7.5}=x:\dfrac{3}{5}\)
\(\Leftrightarrow x:\dfrac{3}{5}=\dfrac{1}{3}\)
hay \(x=\dfrac{1}{5}\)
Bài 2:
a:
1: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
hay \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
Lời giải:
a.
$x:27=-2:3,6=\frac{-5}{9}$
$x=27.\frac{-5}{9}=-15$
b.
$\frac{2x+1}{-27}=\frac{-3}{2x+1}$
$\Rightarrow (2x+1)^2=(-27)(-3)=81=9^2=(-9)^2$
$\Rightarrow 2x+1=9$ hoặc $2x+1=-9$
$\Rightarrow x=4$ hoặc $x=-5$
1.
a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)
b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)
2.
a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)
b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
c) Ta có: \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)
Dấu '=' xảy ra khi x(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
d) Ta có: \(x^2+5y^2-2xy+4y+3\)
\(=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+2\)
\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\forall x,y\)
Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)
a, Ta có: x = 2,5 . 3/5 : 7,5 = 1/5
b, Ta có: x = 3 . 5 : 7 = 15/7
c, Ta có: x = -2 . 37 : 3,6 = -185/9
a) \(2.5\div7.5=x\div\frac{3}{5}\)
\(x\div\frac{3}{5}=\frac{1}{3}\)
\(x=\frac{1}{3}×\frac{3}{5}\)
\(x=\frac{3}{15}=\frac{1}{5}\)
Vậy, \(x=\frac{1}{5}\)
b) \(x\div5=3\div7\)
\(x\div5=\frac{3}{7}\)
\(x=\frac{3}{7}×5\)
\(x=\frac{15}{7}\)
Vậy, \(x=\frac{15}{7}\)
c) \(x\div37=\frac{-2}{3×6}\)
\(x\div37=\frac{-5}{9}\)
\(x=\frac{-5}{9}×37\)
\(x=\frac{-185}{9}\)
Vậy, \(x=\frac{-185}{9}\)
Cbht