`@` `\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

\(4x^2+4x+1=\left(x-2\right)^2\\ \Leftrightarrow\left(2x+1\right)^2=\left(x-2\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=x-2\\2x+1=2-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

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

\(\Leftrightarrow x^2-6x+9+3-x=0\)

\(\Leftrightarrow x^2-7x+12=0\)

\(\Leftrightarrow x^2-3x-4x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Vậy.........

\(\left(x-3\right)^2+3-x=0\\ \Leftrightarrow x^2-6x+9+3-x=0\\ \Leftrightarrow x^2-7x+12=0\\ \Leftrightarrow x^2-3x-4x+12=0\\ \Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Vậy \(S=\left\{3;4\right\}\)

\(\left\{{}\begin{matrix}3x-6y+2z=-4\\3x-y-3z=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3x-6y+2z=-4\\3x-y-3z=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}3x-6y=-4-2z\\3x-y=1+3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5y=1+3z+4+2z\\3x-y=1+3z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5y=5+5z\\3x=y+1+3z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=1+z\\3x=1+z+1+3z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=1+z\\x=\dfrac{4z+6}{3}\end{matrix}\right.\)

\(S=9x^2-8\left(y^2+z^2\right)\)

\(S=9\left(\dfrac{4z+2}{3}\right)^2-8\left[\left(1+z\right)^2+z^2\right]\)

\(S=9.\dfrac{16z^2+16z+4}{9}-8\left[1+2z+z^2+z^2\right]\)

\(S=16z^2+16z+4-8-16z-16z^2\)

\(S=-4\)

Đính chính \(x=\dfrac{4z+2}{3}\) không phải \(x=\dfrac{4z+6}{3}\)