64 x 25 + 35 x 25 + 25 

= ( 64 + 35 + 1 ) x 25

= 100 x 25

= 2500

Giống đề của mk thế

64*25+35*25+25

=64*25+35*25+25*1

=(64+35+1)*25

=100*25

=2500

Câu a đề sai

b

\(Q=y^4+64\)

\(Q=\left(y^4+16y^2+64\right)-16y^2\)

\(Q=\left(y^2+8\right)^2-\left(4y\right)^2\)

\(Q=\left(y^2+8-4y\right)\left(y^2+8+4y\right)\)

lớp mấy bạn ơi

Lớp 6 bạn nhé!

môn anh ak

2x -10 = 0

⇔ 2x = 10

⇔ x= 5

Vậy...

2x - 10 = 0
2x = 0 + 10 
x= 10 : 2
x= 5
 

X = 5

\(\left\{{}\begin{matrix}4x+4y=64\\-2x+5y=10\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}4x+4y=64\\4x-10y=-20\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}14y=64-\left(-20\right)\\x=\dfrac{64-4y}{4}\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}y=6\\x=10\end{matrix}\right.\)

Vậy nghiệm của hệ phương trình (x;y) = (10;6)

\(\left\{{}\begin{matrix}4x+4y=64\\-2x+5y=10\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}4x+4y=64\\-4x+10y=20\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}14y=64\\-4x+10y=20\end{matrix}\right.\)

⇔

`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`

\(\frac{1}{10}+\frac{4}{20}+\frac{9}{30}+.....+\frac{81}{90}\)

\(=\frac{1}{10}+\frac{2}{10}+\frac{3}{10}+...+\frac{9}{10}\)

\(=\frac{\left(9+1\right)\times\left(9+1-1\right):2}{10}\)

\(=\frac{10\times9:2}{10}\)

\(=\frac{45}{10}=4,5\)

=>2x-4x=54-6

=>-2x=48

=>x=24

=>7x-2x=-110

=>5x=110

=>x=22

X(1+2+3+4)=150

10X=150

X=15

 X x ( 1+2+3+4) = 150

 X x 10 = 150

       X = 150 : 10

       X= 15

Vậy X=15

X + 2X + 3X + 4X = 150

          (1+2+3+4)x = 150 

                     10x  = 150

                         x = 15