\(D=5-8x-x^2\\ =-\left[x^2+2.x.4+16\right]+21\\ =-\left(x+4\right)^2+21\le21\forall x\in R\\ \Rightarrow max_D=21.khi.x=-4\)

\(E=4x-x^2+1\\ =-\left(x^2-2.x.2+4^2\right)+17\\ =-\left(x-2\right)^2+17\le17\forall x\in R\\ Vậy:max_E=17.khi.\left(x-2\right)=0\Leftrightarrow x=2\)

\(Sửa,đề:x^2-10x+25\\ =x^2-2x.5+5^2=\left(x-5\right)^2=\left(x-5\right)\left(x-5\right)\\---\\ b,x^3+125=x^3+5^3=\left(x+5\right)\left(x^2-5x+25\right)\\ ---\\ 8x^3-y^3=\left(2x\right)^3-y^3\\ =\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

a, Bạn xem lại đề vì không thể tách được.

b, \(x^3+125\\ =x^3+5^3=\left(x+5\right)\left(x^2-5x+25\right)\)

c, \(8x^3-y^3\\ =\left(2x\right)^3-y^3\\ =\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(a,VP=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ =\left(x+2y\right)\left[x^2-x.2y+\left(2y\right)^2\right]\\ =x^3+\left(2y\right)^3=x^3+8y^3=VT\left(đpcm\right)\\ b,VT=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\left(x-y\right)\\ =x^3-y^3-3xy\left(x-y\right)\\ =x^3-3x^2y+3xy^2-y^3\\ =\left(x-y\right)^3=VP\left(đpcm\right)\)

\(c,VT=\left(x-3y\right)\left(x^2+3xy+9y^2\right)-\left(3y+x\right)\left(9y^2-3xy+x^2\right)\\ =\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]-\left(x+3y\right).\left[x^2-x.3y+\left(3y\right)^2\right]\\ =x^3-27y^3-\left(x^3+27y^3\right)\\ =-54y^3=VP\left(đpcm\right)\)

b, \(x^5+x+1=x^5-x^2+x^2+x+1\\ =x^2\left(x^3-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left[x^2\left(x+1\right)+1\right]\\ =\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)

`@` `\text {Ans}`

`\downarrow`

\(-\dfrac{3}{2}\cdot\left(\dfrac{4}{5}-\dfrac{2}{3}\right)+x=4\left(x-\dfrac{1}{2}\right)\)

\(\Leftrightarrow-\dfrac{3}{2}\cdot\dfrac{2}{15}+x=4x-2\)

\(\Leftrightarrow-\dfrac{1}{5}+x=4x-2\)

\(\Leftrightarrow-\dfrac{1}{5}+2=-x+4x\)

\(\Leftrightarrow\dfrac{9}{5}=3x\)

\(\Leftrightarrow x=\dfrac{9}{5}\div3\)

\(\Leftrightarrow x=\dfrac{3}{5}\)

Vậy, `x=`\(\dfrac{3}{5}\)

-3/2(4/5-2/3)+x=4(x-1/2)

=>-6/5+1+x=4x-2

=>-1/5-2=4x-x

=>-11/5=3x

=>x=-11/5:3=-11/15.

`@` `\text {Ans}`

`\downarrow`

`3.`

\(\left(\dfrac{1}{2}-x\right)+\dfrac{1}{3}=\dfrac{7}{6}-x\)

\(\Leftrightarrow\dfrac{1}{2}-x+\dfrac{1}{3}=\dfrac{7}{6}-x\)

\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{7}{6}=-x+x\)

\(\Leftrightarrow-\dfrac{1}{3}=0\left(\text{Vô lý}\right)\)

Vậy, `x` không có giá trị nào thỏa mãn.

giúp mik với

\(\dfrac{1}{3}\)

giúp mik với ạ

\(x+\left(\dfrac{1}{2}-x\right).\dfrac{1}{3}=\dfrac{7}{6}-x\)

\(\Leftrightarrow x+\dfrac{1}{6}-\dfrac{1}{3}x=\dfrac{7}{6}-x\)

\(\Leftrightarrow2x-\dfrac{1}{3}x=1\)

\(\Leftrightarrow6x-x=3\Leftrightarrow5x=3\Rightarrow x=\dfrac{3}{5}\)

x+(\(\dfrac{1}{2}\)-x)\(\dfrac{1}{3}\)=\(\dfrac{7}{6}\)-x

=>\(\dfrac{1}{6}\)-\(\dfrac{1}{3}\)x=\(\dfrac{7}{6}\)-x-x=\(\dfrac{7}{6}\)-2x

=>\(\dfrac{1}{6}\)-\(\dfrac{7}{6}\)=-2x+\(\dfrac{1}{3}\)x

=>-1=\(\dfrac{-5}{3}\)x

=>1=\(\dfrac{5}{3}\)x

=>x=1:\(\dfrac{5}{3}\)=\(\dfrac{3}{5}\).

Câu a em check lại đề xem mũ 3 hay mũ 4 nhé

mũ 4 chị ơi

KO

Để rút gọn biểu thức, ta sẽ thực hiện các phép tính và kết hợp các thành phần tương tự: P(2x-1).4x^2 + 2x + 1 + (x+1)x^2 - x + 1 = P(8x^3 - 4x^2) + 2x + 1 + x^3 + x^2 - x + 1 = P(8x^3) - P(4x^2) + x^3 + (2x-x) +(1+1) = **8Px^3 - 4Px^2**+ x^3 **+ x**+ **2** Vậy biểu thức đã được rút gọn thành: **8Px³ - 4Px²+x³+x+2**