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

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**

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

\(\Rightarrow\left(3+2x\right)^2-\left(x-5\right)^2=0\)

\(\Rightarrow\left(3+2x+x-5\right)\left(3+2x-x+5\right)=0\)

\(\Rightarrow\left(3x-2\right)\left(x+8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)

b) \(27x^3-54x^2+36x=9\)

\(\Rightarrow27x^3-54x^2+36x-9=0\)

\(\Rightarrow27x^3-54x^2+36x-8+8-9=0\)

\(\Rightarrow\left(3x-2\right)^3-1=0\)

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

\(\Rightarrow\left(3x-3\right)\left[\left(3x-2\right)^2+3x-2+\dfrac{1}{4}-\dfrac{1}{4}+1\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)

\(\Rightarrow\left(3x-3\right)\left[\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\left(1\right)\)

mà \(\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0,\forall x\)

\(\left(1\right)\Rightarrow3x-3=0\Rightarrow3x=3\Rightarrow x=1\)

(\(x-5\))² = (3 +2\(x\))² ⇒ \(\left[{}\begin{matrix}x-5=3+2x\\x-5=-3-2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x\in\){-8; \(\dfrac{2}{3}\)}

  27\(x^3\) - 54\(x^2\) + 36\(x\) = 9

27\(x^3\) - 54\(x^2\) + 36\(x\) - 8 = 1

(3\(x\) - 2)³ = 1 ⇒ 3\(x\) - 2 = 1 ⇒ \(x\) = 1