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23 tháng 4 2023

0,75 x - x + 1 2/5x = 45/100

-0,25x + 1,4x = 45/100

1,15x = 45/100

x = 45/100 : 1,15

x = 9/23

 

7 tháng 5 2020

a)

\(\left(5x+3\right)\cdot\left(x^2+4\right)\cdot\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x+3=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\frac{3}{5}\\x=4\end{matrix}\right.\)

b)

\(\left(4x-1\right)\cdot\left(x-3\right)-\left(x-2\right)\cdot\left(5x+2\right)=0\\ \Leftrightarrow4x^2-12x-x+3-5x^2-2x+10x+4=0\\ \Leftrightarrow-x^2-5x+7=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{5+\sqrt{53}}{2}\\-\frac{5-\sqrt{53}}{2}\end{matrix}\right.\)

c)

\(\left(x+3\right)\cdot\left(x-5\right)+\left(x+3\right)\cdot\left(3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(x-5+3x-4\right)=0\\ \Leftrightarrow\left(x+3\right)\cdot\left(4x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)

d)

\(\left(x+6\right)\cdot\left(3x-1\right)+x^2-36=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x^2-36\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1\right)+\left(x+6\right)\cdot\left(x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(3x-1+x-6\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot\left(4x-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)

e)

\(0.75x\cdot\left(x+5\right)=\left(x+5\right)\cdot\left(3-1.25x\right)\\ \Leftrightarrow0.75x\cdot\left(x+5\right)-\left(x+5\right)\cdot\left(3-1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(0.75x-3+1.25x\right)=0\\ \Leftrightarrow\left(x+5\right)\cdot\left(2x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\frac{3}{2}\end{matrix}\right.\)

a:=>0,75x-x+1,25x=0,2

=>x=0,2

b: =>1/3-x=-3/6+4/6=1/6

=>x=1/3-1/6=1/6

c: =>(x-1)/45=-6/30=-1/5

=>x-1=-9

=>x=-8

d: =>(2/5x-1)=-5/7

=>2/5x=2/7

=>x=5/7

DT
30 tháng 12 2023

\(\dfrac{75}{100}+\dfrac{3}{4}\times29+75\%\times30+0,75\times40\\ =\dfrac{75}{100}\times1+\dfrac{75}{100}\times29+\dfrac{75}{100}\times30+\dfrac{75}{100}\times40\\ =\dfrac{75}{100}\times\left(1+29+30+40\right)\\ =\dfrac{75}{100}\times100=75\)

30 tháng 12 2023

\(\dfrac{75}{100}+\dfrac{3}{4}\) x 29 + 75% x 30 + 0,75 x 40

= 0,75 x 1 + 0,75 x 29 + 0,75 x 30 + 0,75 x 40

= 0,75 x (1 + 29 + 30 + 40)

= 0,75 x 100

= 75

\(A-B=\left(1+\dfrac{3}{4}-\dfrac{3}{4}\right)x^4+\left(-\dfrac{1}{8}x^3+\dfrac{1}{8}x^3\right)+\left(-\dfrac{5}{4}+\dfrac{9}{4}\right)x^2+\left(\dfrac{2}{5}x-\dfrac{2}{5}x\right)+\dfrac{4}{7}+\dfrac{3}{7}\)

\(=x^4+x^2+1>0\forall x\)

19 tháng 12 2021

Answer:

\(x^2-10x+25\)

\(=\left(x-5\right)^2\)

\(=\left(105-5\right)^2\)

\(=10000\)

7 tháng 5 2016

=0,75 x 1  + 0,75 x 29 + 0,75 x 30 + 0,75 x 40

= 0,75 x (1+29+30+40)

=0,75 x 100 

=75 

9 tháng 7 2023

\(-\dfrac{2}{3}\left(x^2y^2\right)^2+\dfrac{1}{3}xy^2\cdot0,75x-3xy\cdot\left(-2x\right)\)

\(=-\dfrac{2}{3}x^4y^4+\dfrac{1}{3}xy^2\cdot\dfrac{3}{4}x-3xy\cdot\left(-2x\right)\)

\(=-\dfrac{2}{3}x^4y^4+\left(\dfrac{1}{3}\cdot\dfrac{3}{4}\right)\left(x\cdot x\right)y^2-\left(3\cdot-2\right)\left(x\cdot x\right)y\)

\(=-\dfrac{2}{3}x^4y^4+\dfrac{1}{4}x^2y^2+6x^2y\)

a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)

\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)

hay \(x\in\left\{30;-30\right\}\)

Vậy: \(x\in\left\{30;-30\right\}\)

b) Ta có: \(\left|x\right|+0.573=2\)

\(\Leftrightarrow\left|x\right|=1.427\)

hay \(x\in\left\{1.427;-1.427\right\}\)

Vậy: \(x\in\left\{1.427;-1.427\right\}\)

c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)

d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)

\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)

\(\Leftrightarrow x=\dfrac{1}{250}\)

Vậy: \(x=\dfrac{1}{250}\)