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a, \(\frac{1}{5}+x=\frac{7}{10}\)

\(x=\frac{7}{10}-\frac{2}{10}=\frac{5}{10}=\frac{1}{2}\)

b, \(\left(\frac{1}{3}-\frac{1}{4}\right):x+25\%=0,2\) 

\(\Leftrightarrow\frac{\frac{1}{3}-\frac{1}{4}}{x}+0,25=0,2\)                             

\(\Leftrightarrow\frac{\frac{1}{12}}{x}=-0,05\)

\(\Leftrightarrow\frac{1}{12}=-0,05x\)

\(\Leftrightarrow x\approx1,7\)

x=7/10-1/5

x=7/10-2/10

x=1/2

(1/3-1/4):x+1/4=1/5

1/12:x+1/4=1/5

1/12:x=1/5+1/4

1/12:x=9/20

x=1/12:9/20

x=1/12×20/9

x=27/5

18 tháng 10 2021

Xin lỗi m.n nhé gửi nhầm tí

a: \(\dfrac{1}{8}\cdot\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{1}{2\cdot3}=\dfrac{1}{6}\)

b: \(=\dfrac{8-3}{12}\cdot\dfrac{6}{5}=\dfrac{6}{12}=\dfrac{1}{2}\)

c: \(=\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{3}{4}\)

d: =63/21+15/21-7/21=71/21

15 tháng 4 2022

\(\dfrac{1}{8}\times\dfrac{4}{5}\times\dfrac{10}{6}=\dfrac{1}{6}\)

\(\left(\dfrac{8}{12}-\dfrac{3}{12}\right)\times\dfrac{6}{5}=\dfrac{5}{12}\times\dfrac{6}{5}=\dfrac{1}{2}\)

\(\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{24}{35}\times\dfrac{35}{32}=\dfrac{3}{4}\)

\(\dfrac{63}{21}+\dfrac{15}{21}-\dfrac{7}{21}=\dfrac{71}{21}\)

`#040911`

`a)`

\(\left(2x-1\right)^2-\left(2x+5\right)\left(2x+1\right)=10\)

\(\Leftrightarrow 4x^2 - 4x + 1 - (4x^2 + 12x + 5) = 10 \\ \Leftrightarrow 4x^2 - 4x + 1 - 4x^2 - 12x - 5 = 10 \\ \Leftrightarrow (4x^2 - 4x^2) - (4x + 12x) + (1 - 5) = 10 \\ \Leftrightarrow -16x - 4 = 10 \Leftrightarrow -16x = 10 + 4 \\ \Leftrightarrow -16x = 14 \\ \Leftrightarrow x = \dfrac{-7}{8}\)

Vậy, `x = -7/8`

`b)`

`9^2(x - 1) + 25(1 - x) = 0`

`<=> 9^2(x - 1) - 25(x - 1) = 0`

`<=> (x - 1)(9^2 - 5^2) = 0`

`<=>`\(\left[{}\begin{matrix}x-1=0\\9^2-5^2=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}x=1\\56=0\left(\text{vô lý}\right)\end{matrix}\right.\)

Vậy, `x = 1`

`c)`

`x^2+3x - 4 = 0`

`<=> x^2 + 4x - x - 4 = 0`

`<=> (x^2 - x) + (4x - 4) = 0`

`<=> x(x - 1) + 4(x - 1) = 0`

`<=> (x + 4)(x - 1) = 0`

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\\ \text{Vậy, }x\in\left\{-4;1\right\}\)

a: =>4x^2-4x+1-(4x^2+2x+10x+5)=10

=>4x^2-4x+1-10-4x^2-12x-5=0

=>-16x-4=0

=>x=-1/4

b: =>(x-1)(9^2-25)=0

=>x-1=0

=>x=1

c: =>x^2+4x-x-4=0

=>(x+4)(x-1)=0

=>x=1 hoặc x=-4

a: \(\dfrac{4}{5}-\dfrac{5}{6}< =\dfrac{x}{30}< =\dfrac{1}{3}-\dfrac{3}{10}\)

=>\(\dfrac{24-25}{30}< =\dfrac{x}{30}< =\dfrac{10-9}{30}\)

=>\(\dfrac{-1}{30}< =\dfrac{x}{30}< =\dfrac{1}{30}\)

=>-1<=x<=1

mà x nguyên

nên \(x\in\left\{-1;0;1\right\}\)

b: \(\dfrac{a}{7}+\dfrac{1}{14}=\dfrac{-1}{b}\)

=>\(\dfrac{2a+1}{14}=\dfrac{-1}{b}\)

=>\(\left(2a+1\right)\cdot b=-14\)

mà 2a+1 lẻ (do a là số nguyên)

nên \(\left(2a+1\right)\cdot b=1\cdot\left(-14\right)=\left(-1\right)\cdot14=7\cdot\left(-2\right)=\left(-7\right)\cdot2\)

=>\(\left(2a+1;b\right)\in\left\{\left(1;-14\right);\left(-1;14\right);\left(7;-2\right);\left(-7;2\right)\right\}\)

=>\(\left(a;b\right)\in\left\{\left(0;-14\right);\left(-1;14\right);\left(3;-2\right);\left(-4;2\right)\right\}\)

20 tháng 1

.

a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)

\(\Leftrightarrow20x-5+2-6x-6x-30=10\)

\(\Leftrightarrow8x=43\)

hay \(x=\dfrac{43}{8}\)

b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)

\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)

\(\Leftrightarrow6x^2-3x-3=0\)

\(\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

9 tháng 9 2021

câu c,d đâu 

a: =>4x^2-24x+36-4x^2+4x-1<10

=>-20x<10-35=-25

=>x>=5/4

b: =>x(x^2-25)-x^3-8<=3

=>x^3-25x-x^3-8<=3

=>-25x<=11

=>x>=-11/25

30 tháng 3 2022
11/12x+3/4=-1/6
20 tháng 2

a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)

= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)

= - 1 + 1  - \(\dfrac{11}{20}\)

=   0 - \(\dfrac{11}{20}\)

= - \(\dfrac{11}{20}\)

b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)

\(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)

\(\dfrac{10}{12}\)

\(\dfrac{5}{6}\)

c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)

= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)

\(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)

\(\dfrac{14}{3}\)