giúp tớ với

( 2m - 3 )( 3n - 2 ) - ( 3m - 2 )( 2n - 3 )

= 6mn - 4m - 9n + 6 - ( 6mn - 9m - 4n + 6 )

= 6mn - 4m - 9n + 6 - 6mn + 9m + 4n - 6

= 5m - 5n

= 5( m - n ) \(⋮\)5 với mọi m, n thuộc Z ( đpcm )

Chịu không hiểu đề, chả biết cái nào có ngoặc cái nào không có ngoặc cả

a, \(3x-5=13\Leftrightarrow3x=18\Leftrightarrow x=6\)

b, \(4x-2=3x+1\Leftrightarrow x=3\)

c, \(5\left(x-3\right)-2\left(x-5\right)=58\Leftrightarrow5x-15-2x+10=58\)

\(\Leftrightarrow3x-5=58\Leftrightarrow3x=63\Leftrightarrow x=21\)

d, \(mx+5x=m^2m^2-25\Leftrightarrow x\left(m+5\right)=m^4-25\)

mk quên phần cuối nhé 

\(\Leftrightarrow x\left(m+5\right)=m^4-25\Leftrightarrow x=\frac{m^4-25}{m+5}\)

1. \(2-\sqrt{\left(3x+1\right)^2}=35\)

<=> \(\left|3x+1\right|=-33\) => pt vô nghiệm

2. \(\sqrt{\left(-2x+1\right)^2}+5=12\)

<=> \(\left|1-2x\right|=12-5\)

<=> \(\left|1-2x\right|=7\)

<=> \(\orbr{\begin{cases}1-2x=7\left(đk:x\le\frac{1}{2}\right)\\2x-1=7\left(đk:x>\frac{1}{2}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}2x=-6\\2x=8\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-3\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)

Vậy S = {-3; 4}

3. ĐKXĐ: \(\sqrt{x^2-1}\ge0\) <=> \(x^2-1\ge0\) <=> \(x^2\ge1\) <=> \(\orbr{\begin{cases}x\ge1\\x\le1\end{cases}}\)

\(\sqrt{x^2-1}+4=0\) <=> \(\sqrt{x^2-1}=-4\)

=> pt vô nghiệm

4. Đk: \(\hept{\begin{cases}\sqrt{5x+7}\ge0\\\sqrt{x+3}>0\end{cases}}\) <=> \(\hept{\begin{cases}5x+7\ge0\\x+3>0\end{cases}}\) <=> \(\hept{\begin{cases}x\ge-\frac{7}{5}\\x>-3\end{cases}}\) => x \(\ge\)-7/5

Ta có: \(\frac{\sqrt{5x+7}}{\sqrt{x+3}}=4\)

<=> \(\left(\frac{\sqrt{5x+7}}{\sqrt{x+3}}\right)^2=16\)

<=> \(\frac{\left(\sqrt{5x+7}\right)^2}{\left(\sqrt{x+3}\right)^2}=16\)

<=> \(\frac{5x+7}{x+3}=16\)

=> \(5x+7=16\left(x+3\right)\)

<=> \(5x+7=16x+48\)

<=> \(5x-16x=48-7\)

<=> \(-11x=41\)

<=> \(x=-\frac{41}{11}\)ktm

=> pt vô nghiệm

1) \(x^4+2x^3-9x^2-10x-24\)

\(=x^4+4x^3+x^2-2x^3-8x^2-2x-2x^2-8x-2\)

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

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

2) \(6x^4+7x^3+5x^2-x-2\)

\(=6x^4-3x^3+10x^3-5x^2+10x^2-5x+4x-2\)

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

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

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

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

3) \(2x^4+3x^3+2x^2-1\)

\(=2x^4+2x^3+x^3+x^2+x^2+x-x-1\)

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

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

4) \(x^3-x^2-x-2\)

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

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

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a) Lập bảng xét dấu

x              0           1              2

x        -     0      +    |       +      |       +

x - 1 -      |       -     0     +      |      +

x - 2 -    |          -    |     -         |      +

Xét các TH xảy ra

TH1: x \(\le\)0 => pt trở thành: -x - 2(1 - x) + 3(2 - x) = 4

<=> - x - 2 + 2x + 6 - 3x = 4 <=> -2x = 4 - 4 <=> -2x = 0 <=> x = 0 (tm)

TH2: 0 < x \(\le\)1 => pt trở thành: x - 2(1 - x) + 3(2 - x) = 4

<=> x - 2 + 2x + 6 - 3x = 4 <=> 4 = 4 (luôn đúng)

TH3: 1 < x \(\le\)2 => pt trở thành: x - 2(x - 1) + 3(2 - x) = 4

<=> x - 2x + 2 + 6 - 3x = 4 <=> -4x = 4 - 8 <=> -4x = -4 <=> x = 1 (ktm)

TH4: x > 2 => pt trở thành: x - 2(x - 1) + 3(x - 2)  = 4

<=> x - 2x + 2 + 3x - 6 = 4 <=> 2x = 4 + 4 <=> 2x = 8 <=> x = 4 (tm)

Vậy ....

Em ko chắc nhé 

a, \(\left|x-1\right|+\left|2-x\right|=3\)

\(\Leftrightarrow\left|x-1+2-x\right|=3\Leftrightarrow\left|1\right|\ne3\)

b, \(\left|x+3\right|+\left|x-5\right|=3x-1\)

\(\Leftrightarrow\left|x+3+x-5\right|=3x-1\)

\(\Leftrightarrow\left|2x-2\right|=3x-1\Leftrightarrow\orbr{\begin{cases}2x-2=3x-1\\-2x+2=3x-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x-1=0\\-5x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{5}\end{cases}}}\)

a)  \(\left|x-1\right|+\left|2-x\right|=3\)

+) TH1 : \(\hept{\begin{cases}x-1\ge0\\2-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le2\end{cases}\Leftrightarrow}1\le x\le2}\)

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

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

\(\Leftrightarrow1=3\)( vô lí ) 

+) TH2 : \(\hept{\begin{cases}x-1\le0\\2-x\le0\end{cases}\Rightarrow\hept{\begin{cases}x\le1\\x\ge2\end{cases}\left(L\right)}}\)

+) TH3 : \(\hept{\begin{cases}x-1\ge0\\2-x\le0\end{cases}\Rightarrow\hept{\begin{cases}x\ge1\\x\ge2\end{cases}\Leftrightarrow}x\ge2}\)

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

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

\(\Leftrightarrow2x-3=3\)

\(\Leftrightarrow x=3\)( Thỏa mãn ) 

+) TH4 : \(\hept{\begin{cases}x-1\le0\\2-x\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\le1\\x\le2\end{cases}\Leftrightarrow}x\le1}\)

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

\(\Leftrightarrow1-x+2-x=3\)

\(\Leftrightarrow3-2x=3\)

\(\Leftrightarrow x=0\) ( thỏa mãn ) 

Vậy tập nghiệm của phương trình là S = { 0 ; 3 }

P/s : ๖²⁴ʱ✰๖ۣۜCɦεɾɾү☠๖ۣۜBσмbʂ✰⁀ᶦᵈᵒᶫッ Ta có : \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\). => Sai rùi nha bạn ^_^

\(x^3+x^2+4=0\)

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

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

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

\(\Leftrightarrow\left(x+2\right)\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{7}{4}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\left(\forall x\right)\end{cases}}\)

\(\Leftrightarrow x=-2\)