a.\(=\dfrac{8}{15}\times\dfrac{1}{2}=\dfrac{4}{15}\)

b.\(=\dfrac{1}{3}\times\dfrac{1}{2}-\dfrac{1}{5}\times\dfrac{1}{2}=\dfrac{1}{6}-\dfrac{1}{10}=\dfrac{1}{15}\)

a.(1/3+1/5) x 1/2

=8/15 x 1/2=8/30=4/15

b)(1/3-1/5) x 1/2=2/15 x 1/2=1/15

\(\left(-x^3.z.y\right).\left(\dfrac{2}{3}.y.x^2\right)^2\)

\(=-x^3.z.y.\dfrac{4}{9}.y^2.x^4\)

\(=-\dfrac{4}{9}x^7.y^3.z\)

`(-x^3zy)(2/3yx^2)^2`

`=-4/9x^3zy.y^2x^4`

`=-4/9x^{3+4}.y^{1+2}z`

`=-4/9x^7y^3z`

\(\left(-x^3zy\right)\left(\dfrac{2}{3}yx^2\right)^2\)

\(\Leftrightarrow\left(-x^3yz\right)\left(\dfrac{2}{3}x^2y\right)\left(\dfrac{2}{3}x^2y\right)\)

\(\Leftrightarrow-\left[x^7y^3z\left(\dfrac{2}{3}\right)^2\right]\)

\(\Leftrightarrow-\dfrac{4}{9}x^7y^3z\)

\(a,=4x^2+4x+1\\ b,=9-12y+4y^2\\ c,=\dfrac{x^2}{4}-xy+y^2\\ d,=\dfrac{25}{4}-5x+x^2\\ e,=4x^2+32xy+64y^2\\ f,=9x^2-30xy+25y^2\)

a. (2x + 1)² 

= 4x² + 4x + 1

b. (3 - 2y)²

= 9 - 12y + 4y²

- Các câu còn lại bn dung CT: (A + B)² = A² + 2AB + B² và (A - B)² = A² - 2AB + B² để tính tiếp nha, phân số cũng đc tính.)

a) 4x^2 + 4x + 1

b) 4y^2 - 12y + 9

c) x^2/4 - xy + y^2

d) x^2 - 5x + 25/4

e) 4x^2 + 32xy + 64y^2

f) 9x^2 - 30xy + 25y^2

bạn nào giúp mình nhớ kết ban voi minh nhe

1/5+8/15:y=1/2

8/15:y=1/2-1/5

8/15:y=3/10

y=8/15:3/10

y=16/9

Vậy y=16/9

8/15:y=1/2-1/5

8/15:y=3/10

y=8/15:3/10

y=16/9

\(\left(-x^2y\right)^3\cdot\dfrac{1}{2}\cdot x^2y^3\cdot\left(-2xy^2z\right)^2\\ =-x^6y^3\cdot\dfrac{1}{2}x^2y^3\cdot4x^2y^4z^2\\ =\left(-1\cdot\dfrac{1}{2}\cdot4\right)\cdot\left(x^6\cdot x^2\cdot x^2\right)\cdot\left(y^3\cdot y^3\cdot y^4\right)\cdot z^2\\ =-2x^{10}y^{10}z^2\)

bạn chụp ảnh sẽ dễ nhìn hơn :D

`(-x^2y)^3. 1/2x^2y^3(-2xy^2z)^2`

`=(-1/2).x^6y^3.x^2y^3.4x^2y^4z^2`

`=(-1/2).4.x^{6+2+2}.y^{3+3+4}.z^2`

`=-2x^10y^10z^2`

`đk:x ne 0,-2`

`a)D=(x/(x+2)+(8x+8)/(x^2+2x)-(x+2)/x):((x^2-x-3)/(x^2+2x)+1/x)`

`=((x^2+8x+8-x^2-4x-4)/(x(x+2))):((x^2-x-3+x+2)/(x(x+2)))`

`=(4x+4)/(x(x+2)):(x^2-1)/(x(x+2))`

`=(4x+4)/(x^2-1)(x ne +-1)`

`=4/(x-1)`

`b)x(x-2)-(x-2)=0`

`<=>(x-2)(x-1)=0`

Vì `x ne 1=>x-1 ne 0`

`=>x-2=0<=>x=2`

`=>D=4/(2-1)=4`

`c)D<0`

Mà `4>0`

`=>x-1<0`

`=>x<1`

Kết hợp đkxđ:

`=>x<1,x ne 0,x ne -2`

`d)D=2`

`<=>4/(x-1)=2`

`<=>2/(x-1)=1`

`<=>x-1=2`

`<=>x=3(tm)`

Câu 2 bạn ghi thiếu đề

Câu 1:

\(\Leftrightarrow\left(m^2-3m\right)x+2x< 2-m\)

\(\Leftrightarrow\left(m^2-3m+2\right)x< 2-m\)

BPT đã cho vô nghiệm khi và chỉ khi:

\(\left\{{}\begin{matrix}m^2-3m+2=0\\2-m\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=1\\m=2\end{matrix}\right.\\m\ge2\end{matrix}\right.\) \(\Rightarrow m=2\)

\(\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}=\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}\) \(\left(dkxd:x\ne-2\right)\)

\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)

Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm

\(\Leftrightarrow6x+2-x=0\)

\(\Leftrightarrow5x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)

Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)

\(\dfrac{\left(x^2+x+1\right).\left(3x+1\right)}{x+2}=\dfrac{\left(x^2+x+1\right).x}{2\left(x+2\right)}\)

hay \(\left(x^2+x+1\right).\dfrac{3x+1}{x+2}=\left(x^2+x+1\right).\dfrac{x}{2\left(x+2\right)}\)

Đề bài yêu cầu gì?

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\Rightarrow x=2k;y=3k\)

\(T=\dfrac{2x^2-y^2}{2x^2+y^2}=\dfrac{2\left(2k\right)^2-\left(3k\right)^2}{2\left(2k\right)^2+\left(3k\right)^2}=\dfrac{8k^2-9k^2}{8k^2+9k^2}=\dfrac{-k^2}{17k^2}=\dfrac{-1}{17}\)