`y xx1,7+y xx2,1=9,12`

`y xx(1,7+2,1)=9,12`

`y xx3,8=9,12`

`y=9,12:3,8`

`y=2,4`

\(y\times1,7+y\times2,1=9,12\)

\(y\times\left(1,7+2,1\right)=9,12\)

\(y\times3,8=9,12\)

\(y=9,12:3,8\)

\(y=2,4\)

Yx3,8=9,12

Y=2,4

A   6 x 8 = 48

 Vậy Y bằng 56 : 8, y = 7

B   Ko biết

1)

dãy trên có số chữ số là:

( 26 - 4 ) : 2 + 1 = 12 chữ số

tổng là:

( 26 + 4 ) x 12 : 2 = 180

ta có:

x + 180 = 210

x          = 210 - 180

x          = 30

a) \(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30\)

\(\Rightarrow15x-9x^2-25+15x+9\left(x^2+2x+1\right)-30=0\)

\(\Rightarrow30x-9x^2-25+9x^2+18x+9-30=0\)

\(\Rightarrow48x-46=0\)

\(\Rightarrow x=\frac{23}{24}\)

b) \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Rightarrow\left(x^2+8x+16\right)-\left(x^2-1\right)=16\)

\(\Rightarrow x^2+8x+16-x^2+1=16\)

\(\Rightarrow8x+17=16\)

\(\Rightarrow8x=-1\)

\(\Rightarrow x=\frac{-1}{8}\)

c) \(\left(y-2\right)^3-\left(y-3\right)\left(y^2+3y+9\right)+6\left(y+1\right)^2=49\)

\(\Rightarrow\left(y-2\right)^3-\left(y^3-3^3\right)+6\left(y^2+2y+1\right)=49\)

\(\Rightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\)

\(\Rightarrow\left(y^3-y^3\right)+\left(-6y^2+6y^2\right)+\left(12y+12y\right)+\left(-8+27+6\right)=49\)

\(\Rightarrow24y+25=49\)

\(\Rightarrow24y=24\)

\(\Rightarrow y=1\)

d) \(\left(y+3\right)^3-\left(y+1\right)^3=56\)

\(\Rightarrow\left(y+3-y-1\right)[\left(y+3\right)^2+\left(y+3\right)\left(y+1\right)+\left(y+1\right)^2]=56\)

\(\Rightarrow2\left(y^2+6y+9+y^2+4y+3+y^2+2y+1\right)=56\)

\(\Rightarrow3y^2+12y+13=28\)

\(\Rightarrow\left(3y^2+15y\right)-\left(3y+15\right)=0\)

\(\Rightarrow3y\left(y+5\right)-3\left(y+5\right)=0\)

\(\Rightarrow3\left(y-1\right)\left(y+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

a) 15 x y - 2009 = 1201

 \(15.x.y=1201+2009\)

\(15.x.y=3210\)

\(x.y=3210:15\)

\(x.y=214\)

Tới đây bó tay

b)[(z + 32) - 15 ] . 2 = 60

\(\left(z+32\right)-15=60:2\)

\(\left(z+32\right)-15=30\)

\(\left(z+32\right)=30+15\)

\(z+32=45\)

\(z=45-32\)

\(z=13\)

c)[161 + (153-z)] .18 = 3870

\(161+\left(153-z\right)=3870:18\)

\(161+\left(153-z\right)=215\)

\(153-z=215-161\)

\(153-z=54\)

\(-z=54-153\)

\(-z=-99\)

\(z=99\)

d)[43 - (56 - z)] . 12 = 384

\(43-\left(56-z\right)=384:12\)

\(43-\left(56-z\right)=32\)

\(-\left(56-z\right)=32-43\)

\(56+z=-11\)

\(z=-11-56\)

\(z=-11-56\)

\(z=-67\)

Mình có thể sai đâu đó

Mong bạn thông cảm

Chúc bạn học giỏi

Ap dụng tính chất DTSBN ta có

\(\frac{x}{9}=\frac{y}{3}=\frac{z}{8}=\frac{x-y+x}{9-3+8}=\frac{56}{14}=4\)

\(+\frac{x}{9}=4=>x=36\)

\(+\frac{y}{3}=4=>y=12\)

\(+\frac{z}{8}=4=>z=32\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có

 \(\frac{x}{9}=\frac{y}{3}=\frac{z}{8}=\frac{x-y+z}{9-3+8}=\frac{56}{14}=4\)

=> \(\hept{\begin{cases}\frac{x}{9}=4\\\frac{y}{3}=4\\\frac{z}{8}=4\end{cases}}\) => \(\hept{\begin{cases}x=4.9=36\\y=4.3=12\\z=4.8=32\end{cases}}\)

Vậy ....

cảm ơn mấy bạn!

Theo đề ra ta có :  \(\frac{x}{2,5}\),\(\frac{y}{4}\)\(\frac{z}{1,6}\)và 4.x - 8.y + 5.z =-56

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

\(\frac{x}{2,5}\)=\(\frac{y}{4}\)=\(\frac{z}{1,6}\)= \(\frac{4x}{10}\)=\(\frac{8y}{32}\)=\(\frac{5z}{8}\)=\(\frac{4x-8y+5z}{10-32+8}\)=\(\frac{-56}{-14}=4\)

=> \(\hept{\begin{cases}\frac{x}{2,5}=4\Rightarrow x=10\\\frac{y}{4}=4\Rightarrow y=16\\\frac{z}{1,6}=4\Rightarrow z-6,4\end{cases}}\)