\(\dfrac{7}{5}+\dfrac{4}{7}-\dfrac{9}{10}=\dfrac{49-20}{35}-\dfrac{9}{10}=\dfrac{19}{35}-\dfrac{9}{10}=\dfrac{190-315}{350}=\dfrac{-125}{350}\)

\(\dfrac{2}{1}+\dfrac{3}{4}\text{×}\dfrac{8}{5}=\dfrac{8+3}{4}\text{×}\dfrac{8}{5}=\dfrac{11\text{×}8}{4\text{×}5}=\dfrac{88}{20}\)

mấy câu kia áp dụng là dc!

x.10=(x-10).30

=>10x=30x-300

=>10x-30x+300=0

=>-20x-300=0

=>x=15

\(\Leftrightarrow\dfrac{x}{30}-\dfrac{x-10}{10}=0\)

\(\Leftrightarrow\dfrac{x-3\left(x-10\right)}{30}=0\)

\(\Leftrightarrow x-3x+30=0\)

\(\Leftrightarrow-2x+30=0\)

\(\Leftrightarrow-2x=-30\)

\(\Leftrightarrow x=15\)

\(=>x.10=30x-300\)

\(=>10x-30x-300=0\)

\(=>-20x=300\)

\(=>x=-15\)

`2x+5y=11(1)`

`2x-3y=0(2)`

Lấy (1) trừ (2)

`=>8y=11`

`<=>y=11/8`

`<=>x=(3y)/2=33/16`

a) Ta có: \(\left\{{}\begin{matrix}2x+5y=11\\2x-3y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8y=11\\2x-3y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{11}{8}\\2x=3y=3\cdot\dfrac{11}{8}=\dfrac{33}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{33}{16}\\y=\dfrac{11}{8}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{33}{16}\\y=\dfrac{11}{8}\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}4x+3y=6\\2x+y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+3y=6\\4x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\2x+y=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-2=4\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=6\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(3;-2)

`b)4x+3y=6(1)`

`2x+y=4<=>4x+2y=8(2)`

Lấy (1) trừ (2) ta có:

`y=-2`

`<=>x=(4-2y)/2=3`

`-2/5 : x=1/2`

`=> x= -2/5 : 1/2`

`=> x= -2/5 xx 2`

`=>x= -4/5`

__

`7/6 : x = 7/4`

`=>x= 7/6 : 7/4`

`=>x=7/6 xx 4/7`

`=>x= 28/42`

`=>x=2/3`

tìm x hả

\(x=\dfrac{-2}{5}:\dfrac{1}{2}\)

\(x=\dfrac{-4}{5}\)

b. \(x=\dfrac{7}{6}:\dfrac{7}{4}\)

\(x=\dfrac{2}{3}\)

Ta có : \(x^2+x+4=x^2+x+\frac{1}{4}+\frac{15}{4}=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\left(\forall x\right)\)

+) \(\left(x-1\right)\left(x^2+x+4\right)=0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(\left(x-1\right)\left(x^2+x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2+x=-4\end{cases}}\)

+) x² + x = - 4

<=> ( x + 1/2 )² = - 4 + 1/4 = -15/4

Mà ( x + 1/2 )² lớn hơn hoặc bằng 0 với mọi x

=> x² + x + 4 = 0 ktm

Vậy pt = 0 <=> x = 1

\(C=\dfrac{-5}{7}+\dfrac{-2}{7}+\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{-1}{5}=-1+1-\dfrac{1}{5}=\dfrac{-1}{5}\)

\(\dfrac{2}{5}\cdot\dfrac{3}{4}+1\dfrac{3}{10}=\dfrac{3}{10}+\dfrac{13}{10}=\dfrac{16}{10}=\dfrac{8}{5}\)

2/5*3/4+1/10 =6/20 +6/20 =12/20=3/5

8/5

a)Ta có:

 \(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)

\(\Rightarrow x-3,5=y-\dfrac{1}{10}=0\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}=0,1\end{matrix}\right.\)

b) Ta có:

\(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)

b: ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)