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) \(x^3-x^2-x-2\)

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

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

\(\frac{-27}{5}.\frac{-4}{15}+\frac{-4}{15}.\left(2010+\frac{27}{5}\right)\)

\(=\frac{-4}{15}.\left(\frac{-27}{5}+2010+\frac{27}{5}\right)\)

\(=\frac{-4}{15}.2010\)

\(=-536\)

Học tốt

\(-\frac{27}{5}.\left(-\frac{4}{15}\right)+\left(-\frac{4}{15}\right).\left(2010-\frac{-27}{5}\right)\)

\(=\left(-\frac{4}{15}\right)\left(2010-\frac{-27}{5}+\frac{-27}{5}\right)\)

\(=-\frac{4}{15}.\left(2010+\frac{27}{15}-\frac{27}{15}\right)\)

\(=-\frac{4}{15}.2010=-536\)

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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 ....

Đổ i \(2\begin{cases}3\\5\end{cases}\) m= 26dm

Sợi dây thứ 2 dài số m là : 26-19=7(dm) = 0,7m

                                             Đáp số 0,7m

cho mình hỏi 0,7m là gì, mình chưa học?

Giả sử \(\frac{a}{b}=\frac{c}{d}\)Suy ra  điều ta cần chứng minh là \(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}\)

Theo tính chất của dãy tỉ số bằng nhau :

\(\hept{\begin{cases}\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\\\frac{a}{b}=\frac{3c}{3d}=\frac{a+3c}{b+3d}\end{cases}}< =>\frac{a+3c}{b+3d}=\frac{a+c}{b+d}\)

Vậy ta có điều phải chứng minh

Ta có : \(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}\)

=> (a + 3c)(b+ d) = (b + 3d)(a + c)

=> ab +ad + 3bc + 3cd = ab + bc + 3ad + 3cd

=> ad + 3bc  = bc + 3ad

=> 3bc - bc = 3ad - ad

=> 2bc = 2ad

=> bc = ad

=> \(\frac{a}{b}=\frac{c}{d}\) (đpcm)

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 ^_^

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

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

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

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

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

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

\(\Leftrightarrow x=-2\)