(|-4/7|+2020/2021) : 2/9 + (1/2021+3/7):2/9

=(4/7+2020/2021) : 2/9 + (1/2021 + 3/7) : 2/9

=(4/7 + 2020/2021 + 1/2021 + 3/7) : 2/9

=[(4/7 + 3/7) + (2020/2021 + 1/2021)] : 2/9

=[ 1 + 1 ] : 2/9 = 2 : 2/9 =2 . 9/2 = 9

\(\frac{x}{6}=\frac{z}{5}\Leftrightarrow\frac{x}{12}=\frac{z}{10}\) (*)

\(\frac{y}{4}=\frac{z}{2}\Leftrightarrow\frac{y}{20}=\frac{z}{10}\) (**)

Tu (*) va (**) \(\Leftrightarrow\frac{x}{12}=\frac{y}{20}=\frac{z}{10}\)

Ap dung tinh chat day ty so bang nhau

\(\frac{x}{12}=\frac{y}{20}=\frac{z}{10}=\frac{2x+3y-z}{2.12+3.20-10}=\frac{-148}{74}=-2\)

\(\Leftrightarrow\frac{x}{12}=-2\Leftrightarrow x=\left(-2\right).12=-24\)

\(\Leftrightarrow\frac{y}{20}=-2\Leftrightarrow y=\left(-2\right).20=-40\)

\(\Leftrightarrow\frac{z}{10}=-2\Leftrightarrow z=\left(-2\right).10=-20\)

theo t/c day ti số = nhau ta có:

12x-15y/7=20z-12x/9=15y-20z/11=12x-15y+20z-12x15y-20z/7+9+11=0

suy ra 12x-15y=0, 20z-12x=0

suy ra 12x=15y,20z=12x

suy ra 4/5x=y,z=3/5x (1)

thay (1) vào x+y+z=48 :

4/5x+3/5x+x=48

x.12/5=48

x=20

thay x=20 vào (1):

y=4/5.20 suy ra y=16

z=3/5.20 suy ra z=12

vậy :...

Ta có: \(\frac{\left(12x-15y\right)}{7}=\frac{\left(20z-12x\right)}{9}=\frac{\left(15y-20z\right)}{11}\)\(=0\)(đoạn này là bn cộng tử lại nha)

=>12x=15y

20z=12x

15y=20z

=> 12x=15y=20z

=>y=4/5x;z=3/5x

Thay vào x+y+x=48, ta có: 12/5x=48

=> x=20, y=16, z=12

\(\frac{x}{6}=\frac{z}{5}\Leftrightarrow\frac{x}{12}=\frac{z}{10},\frac{y}{4}=\frac{z}{2}\Leftrightarrow\frac{y}{20}=\frac{z}{10}\)

suy ra \(\frac{x}{12}=\frac{y}{20}=\frac{z}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có: 

\(\frac{x}{12}=\frac{y}{20}=\frac{z}{10}=\frac{2x+3y-z}{2.12+3.20-10}=\frac{-148}{74}=-2\)

\(\Leftrightarrow\hept{\begin{cases}x=-2.12=-24\\y=-2.20=-40\\z=-2.10=-20\end{cases}}\)

\(A=x^3.\left(\frac{-5}{4}\right)x^2y.\left(\frac{2}{5}x^3y^4\right)\)

\(A=\left(-\frac{5}{4}.\frac{2}{5}\right).\left(x^3.x^2.x^3\right).\left(y.y^4\right)\)

\(A=-\frac{1}{2}x^8y^5\)

\(|x^2-2|+\sqrt{y^2-4y\sqrt{2}+8}+|2x+y|=0\)

Do \(|x^2-2|\ge0;\sqrt{y^2-4y\sqrt{2}+8}\ge0;|2x+y|\ge0\)

\(=>|x^2-2|+\sqrt{y^2-4y\sqrt{2}+8}+|2x+y|\ge0\)

Mà \(|x^2-2|+\sqrt{y^2-4y\sqrt{2}+8}+|2x+y|=0\)

Nên dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}x^2-2=0\\y^2-4y\sqrt{2}+8=0\\2x+y=0\end{cases}}< =>\hept{\begin{cases}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(y-2\sqrt{2}\right)^2=0\\2x+y=0\end{cases}}\)

\(< =>\hept{\begin{cases}x=\sqrt{2}orx=-\sqrt{2}\\y=2\sqrt{2}\\2x+2\sqrt{2}=0\end{cases}}\)\(< =>\hept{\begin{cases}x=-\sqrt{2}\\y=2\sqrt{2}\end{cases}}\)

Vậy .... 

a) \(3x^2-2x\left(5+1,5x\right)+10=3x^2-10x-3x^2+10=10-10x\)

b) \(7y\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3,5x\right)\)

\(=28y^2-7xy+4y^2-28xy-4y^2+7x=28y^2-35xy+7x\)

c) \(\left\{2x-3\left(x-1\right)-5\left[x-4\left(3-2x\right)+10\right]\right\}\left(-2x\right)\)

\(=\left\{2x-3x+3-5\left[x-12+8x+10\right]\right\}\left(-2x\right)\)

\(=\left(3-x-5x+12-8x-50\right)\left(-2x\right)=\left(-35-14x\right)\left(-2x\right)\)

\(=70x+28x^2\)

tiền đâu cho đi rùi làm

| -5x | = 5/3 

x=-1/3, x=1/3

| 2x-1|=1/4

x=3/8, x=5/8

mới lớp 5 nên ko biết câu này hihi