a: \(x\left(1-2x\right)+2x^2=14\)

=>\(x-2x^2+2x^2=14\)

=>x=14

b: \(x\left(x-5\right)+3x-15=0\)

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

=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

(2x - 1)³ - 8x + 4 = 0

(2x - 1)³ - 4x(2x - 1) = 0

(2x - 1)[(2x - 1)² - 4x] = 0

(2x - 1)[(2x - 1)(2x - 1) - 4x] = 0

(2x - 1)[2x(2x - 1) - 1.(2x - 1) - 4x] = 0

(2x - 1)(4x² - 2x - 2x + 1 - 4x) = 0

(2x - 1)(4x² + 1) = 0

⇒ 2x - 1 = 0 hoặc 4x² + 1 = 0

*) 2x - 1 = 0

2x = 1

x = 1/2

*) 4x² + 1 = 0

4x² = -1 (vô lý vì 4x² ≥ 0 với mọi x)

Vậy x = 1/2

`(2x-1)^3-8x+4=0`

`-> (2x-1)^3 - 4(2x-1)=0`

`-> (2x-1)^2*(2x-1)-4(2x-1)=0`

`-> (2x-1)* [(2x-1)^2-4]=0`

`->` TH1: `2x-1=0`

`-> 2x=1`

`-> x=1/2`

TH2: `(2x-1)^2-4=0`

`-> (2x-1)^2=4`

`-> (2x-1)^2=(+-2)^2`

`->`\(\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\)

`->`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy, `x={1/2; 3/2; -1/2}`.

\(a)\left(x-2\right)\left(x^2+2x-3\right)\ge0.\)

Đặt \(f\left(x\right)=\left(x-2\right)\left(x^2+2x-3\right).\)

Ta có: \(x-2=0.\Leftrightarrow x=2.\\ x^2+2x-3=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-3.\end{matrix}\right.\)

Bảng xét dấu:

x                   \(-\infty\)       -3       1       2     \(+\infty\)

\(x-2\)                    -      |    -   |   -   0   +

\(x^2+2x-3\)         +     0    -   0  +   |    +

\(f\left(x\right)\)                     -     0    +  0   -  0   +

Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left[-3;1\right]\cup[2;+\infty).\)

\(b)\dfrac{x^2-9}{-x+5}< 0.\)

Đặt \(g\left(x\right)=\dfrac{x^2-9}{-x+5}.\)

Ta có: \(x^2-9=0.\Leftrightarrow\left[{}\begin{matrix}x=3.\\x=-3.\end{matrix}\right.\)

\(-x+5=0.\Leftrightarrow x=5.\)

Bảng xét dấu:

x            \(-\infty\)      -3       3        5       \(+\infty\)

\(x^2-9\)            +   0   -   0   +   |    +

\(-x+5\)          +    |   +   |    +  0    -

\(g\left(x\right)\)              +    0   -   0   +  ||    -

Vậy \(g\left(x\right)< 0.\Leftrightarrow x\in\left(-3;3\right)\cup\left(5;+\infty\right).\)

1: Ta có: \(\left(3-x\right)^2+\left(2x+1\right)^2-\left(2-x\right)^2-\left(2x+1\right)^2=0\)

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

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

\(\Leftrightarrow x=\dfrac{5}{2}\)

2: Ta có: \(\left(1-2x\right)^2-3\left(x-1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow4x^2-4x+1-3x^2+6x-3+\left(x+1\right)^2-2\left(x-1\right)^2=0\)

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

\(\Leftrightarrow2x^2+4x+1-2x^2+4x-2=0\)

\(\Leftrightarrow x=\dfrac{1}{8}\)

a: f(-2)=4+3=7

f(-1)=2+3=5

f(0)=3

f(1/2)=-1+3=2

f(-1/2)=1+3=4

b: g(-1)=1-1=0

f(0)=0-1=-1

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

\(\Leftrightarrow x=0\) hay \(x+3=0\) hay \(x^2+1=0\) (pt vô nghiệm vì \(x^2+1\ge1\))

\(\Leftrightarrow x=0\) hay \(x=-3\)

-Vậy \(S= \left\{0;-3\right\}\)

Ta có: \(\left(1-x\right)^2+\left(x-x^2\right)+3=0\)

\(\Leftrightarrow x^2-2x+1+x-x^2+3=0\)

\(\Leftrightarrow4-x=0\)

hay x=4

Vậy: S={4}

$⇔x^2-2x+1+x-x^2+3=0$

$⇔-x=-4$

$⇔x=4$

Vậy phương trình đã cho có tập nghiệm S={4}