a) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)

\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)

\(\Leftrightarrow-14x=14\)

\(\Leftrightarrow x=-1\)

b) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(\Leftrightarrow17x=-34\Rightarrow x=-2\)

`@` `\text {Ans}`

`\downarrow`

`3,8 * 2x = 1/4*8/3`

`=> 3,8*2x = 2/3`

`=> 2x = 2/3 \div 3,8`

`=> 2x = 10/57`

`=> x = 10/57 \div 2`

`=> x = 5/57`

Vậy, `x = 5/57.`

Mình làm lại theo yc nhé

`3,8 \div 2x = 1/4 \div 8/3`

`=> 3,8*8/3 = 2x*1/4`

`=> 152/15 = 2x*1/4`

`=> 2x = 152/15 \div 1/4`

`=> 2x = 608/15`

`=> x = 608/15 \div 2`

`=> x = 304/15`

Vậy, `x = 304/15.`

Bài 1:

$x-1=|2x-1|\geq 0\Rightarrow x\geq 1$

$\Rightarrow 2x-1>0\Rightarrow |2x-1|=2x-1$. Khi đó:

$2x-1=x-1\Leftrightarrow x=0$  (không thỏa mãn vì $x\geq 1$)

Vậy không tồn tại $x$ thỏa đề.

Bài 2:

Nếu $x\geq \frac{1}{3}$ thì:

$3x-1=2x+3$

$\Leftrightarrow x=4$ (tm)

Nếu $x< \frac{1}{3}$ thì:

$1-3x=2x+3$

$\Leftrightarrow -2=5x\Leftrightarrow x=\frac{-2}{5}$ (tm)

Vậy......

Bài 3: Xét các TH sau:

TH1: $x\geq 2$ thì:

$x-1+x-2=3$

$2x-3=3$

$2x=6$

$x=3$ (thỏa mãn)

TH2: $1\leq x< 2$ thì:

$x-1+2-x=3$

$1=3$ (vô lý- loại)

TH3: $x< 1$

$1-x+2-x=3$

$3-2x=3$

$2x=0$

$x=0$ (thỏa mãn)

Giải pt à bạn

Bài 1:

a) \(x\left(x+1\right)+x\left(x-1\right)-2x^2\)

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

\(=2x^2-2x^2\)

\(=0\)

b) \(\left(x+2\right)\left(x^2-x+1\right)-\left(x-2\right)\left(x^2+x+1\right)\)

\(=x^3-x^2+x+2x^2-2x+2-x^3-x^2-x+2x^2+2x+2\)

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

\(=2x^2+4\)

c) \(\left(3-x\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)

\(=\left(x-3\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)

\(=\left[\left(x-3\right)+\left(x+7\right)\right]^2\)

\(=\left(x-3+x+7\right)^2\)

\(=\left(2x+4\right)^2\)

Bài 2:

a) \(x^2-4-4\left(2-x\right)\)

\(=\left(x^2-4\right)+4\left(x-2\right)\)

\(=\left(x^2-2^2\right)+4\left(x-2\right)\)

\(=\left(x+2\right)\left(x-2\right)+4\left(x-2\right)\)

\(=\left(x-2\right)\left(x+2+4\right)\)

\(=\left(x-2\right)\left(x+6\right)\)

b) \(x^3+8\left(1-2x\right)^3\)

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

\(=\left[x+2\left(1-2x\right)\right]\left[x^2-x\cdot2\left(1-2x\right)+2^2\left(1-2x\right)^2\right]\)

\(=\left(x+2-4x\right)\left[x^2-2x\left(1-2x\right)+4\left(1-4x+4x^2\right)\right]\)

\(=\left(2-3x\right)\left(x^2-2x+4x^2+4-16x+16x^2\right)\)

\(=\left(2-3x\right)\left(21x^2-18x+4\right)\)

c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)

Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)

Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)

\(\left(x+3\right)^2-\left(4-x\right)\left(4+x\right)=10\)

<=>   \(x^2+6x+9-\left(16-x^2\right)=10\)

<=>   \(2x^2+6x-17=0\)

<=>  \(x^2+3x-\frac{17}{2}=0\)

<=>  \(\left(x+\frac{3}{2}\right)^2-\frac{43}{4}=0\)

<=>  \(\left(x+\frac{3}{2}+\frac{\sqrt{43}}{2}\right)\left(x+\frac{3}{2}-\frac{\sqrt{43}}{2}\right)=0\)

<=>  \(\orbr{\begin{cases}x+\frac{3}{2}+\frac{\sqrt{43}}{2}=0\\x+\frac{3}{2}-\frac{\sqrt{43}}{2}=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=\frac{-3-\sqrt{43}}{2}\\x=\frac{\sqrt{43}-3}{2}\end{cases}}\)

Vậy...

\((x+3)^2-(4-x)(4+x)=10\)

\(\Rightarrow x^2+6x+9-(16+4x-4x+x^2)=10\)

\(\Rightarrow x^2+6x+9-16-x^2=10\)

\(\Rightarrow6x+9=26\)

\(\Rightarrow6x=17\)

\(\Rightarrow x\in\varnothing\)

C

C

c?

a, 3x - 2x < 6 <=> x < 6 

b, đk : x khác -1 ; 3 

=> x^2 - 3x = x^2 - x - 2 

<=> -2x = -2 <=> x = 1 (tm)