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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 ^_^

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

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

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

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

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

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

\(\Leftrightarrow x=-2\)

a) Bảng xét dấu:

\(\Rightarrow\left(3x-9\right)\left(2x+4\right)< 0\Leftrightarrow-2< x< 3\)

a) ( 2x + 4 )( 3x - 9 ) < 0

Xét hai trường hợp :

1/ \(\hept{\begin{cases}2x+4< 0\\3x-9>0\end{cases}}\Rightarrow\hept{\begin{cases}2x< -4\\3x>9\end{cases}}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\)( loại )

2/ \(\hept{\begin{cases}2x+4>0\\3x-9< 0\end{cases}}\Rightarrow\hept{\begin{cases}2x>-4\\3x< 9\end{cases}}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}}\Rightarrow-2< x< 3\)

Vậy tập nghiệm của bất phương trình là -2 < x < 3

b) \(\frac{x^2+5}{x-5}>0\)

Rõ ràng \(x^2+5>0\forall x\)

=> Để \(\frac{x^2+5}{x-5}>0\)

=> x - 5 > 0

=> x > 5

Vậy tập nghiệm của bất phương trình là x > 5

c) x² - 15x + 50 \(\ge\)0

<=> x² - 5x - 10x + 50 \(\ge\)0

<=> x( x - 5 ) - 10( x - 5 ) \(\ge\)0

<=> ( x - 10 )( x - 5 ) \(\ge\)0 

Xét 2 trường hợp

1/ \(\hept{\begin{cases}x-10\ge0\\x-5\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge10\\x\ge5\end{cases}}\Rightarrow x\ge10\)

2/ \(\hept{\begin{cases}x-10\le0\\x-5\le0\end{cases}}\Rightarrow\hept{\begin{cases}x\le10\\x\le5\end{cases}}\Rightarrow x\le5\)

Vậy tập nghiệm của bất phương trình là \(x\le5\)hoặc \(x\ge10\)

d) x² - 6x + 15 > 0

<=> x² - 6x + 9 + 6 > 0

<=> ( x - 3 )² + 6 > 0 ( đúng với mọi x )

Vậy bất phương trình có vô số nghiệm 

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

\(=20x^2-20x+5+4x^2+12x-4x-12-50+60x-18x^2\)

\(=6x^2+48x-57\)

b, \(\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)\)

\(=81x^2-18x+1+1-10x+25x^2+18x-90x^2-2+10x\)

\(=16x^2\)

c;d;e;f tự làm, đầu I giữ lấy còn trường tồn:) 

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

\(=5\left(4x^2-4x+1\right)+4\left(x^2+2x-3\right)-2\left(25-30x+9x^2\right)\)

\(=20x^2-20x+5+4x^2+8x-12-50+60x-18x^2\)

\(=\left(20x^2+4x^2-18x^2\right)+\left(60x+8x-20x\right)+\left(5-12-50\right)\)

\(=6x^2+48x-57\)

a) 

\(=\frac{8^3}{\left(-8\right)^{-5}}=\frac{8^3}{-\frac{1}{8^5}}=8^3.-\left(8\right)^5=-8^8\)

b)

\(=\frac{15x^2y^2}{5xy^2}=3x\)

khó thế