TÌM X : \(2\left(x-1\right)+3\left(x-2\right)=x-4\)
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a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
1: Ta có: \(4x^2-36=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)
\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)
\(\Leftrightarrow2x=10\)
hay x=5
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
\(\left|x-1\right|+2\left|x-2\right|+3\left|x-3\right|+4\left|x-4\right|+5\left|x-5\right|+20x=0\left(1\right)\)
TH1: x<1
(1) trở thành 1-x+2(2-x)+3(3-x)+4(4-x)+5(5-x)+20x=0
=>\(1-x+4-2x+9-3x+16-4x+25-5x+20x=0\)
=>\(5x+55=0\)
=>x=-11(nhận)
TH2: 1<=x<2
Phương trình (1) sẽ trở thành:
\(x-1+2\left(2-x\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+4-2x+9-3x+16-4x+25-5x+20x=0\)
=>\(7x+53=0\)
=>\(x=-\dfrac{53}{7}\left(loại\right)\)
TH3: 2<=x<3
Phương trình (1) sẽ trở thành:
\(x-1+2\left(x-2\right)+3\left(3-x\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+9-3x+16-4x+25-5x+20x=0\)
=>\(11x+45=0\)
=>\(x=-\dfrac{45}{11}\left(loại\right)\)
TH4: 3<=x<4
Phương trình (1) sẽ trở thành:
\(x-1+2\left(x-2\right)+3\left(x-3\right)+4\left(4-x\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+3x-9+16-4x+25-5x+20x=0\)
=>\(-3x+27=0\)
=>x=9(loại)
TH5: 4<=x<5
Phương trình (1) sẽ trở thành:
\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(5-x\right)+20x=0\)
=>\(x-1+2x-4+3x-9+4x-16+25-5x+20x=0\)
=>\(25x-5=0\)
=>x=1/5(loại)
TH6: x>=5
Phương trình (1) sẽ trở thành:
\(\left(x-1\right)+2\left(x-2\right)+3\left(x-3\right)+4\left(x-4\right)+5\left(x-5\right)+20x=0\)
=>\(x-1+2x-4+3x-9+4x-16+5x-25+20x=0\)
=>35x-55=0
=>x=55/35(loại)
\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}\left(x^2+4x+4\right)-\dfrac{5}{4}\left(x^2-1\right)=\dfrac{3}{2}x\left(x-2\right)-x-4\)
\(\Leftrightarrow\dfrac{1}{2}x^2-3x-\dfrac{9}{2}-\dfrac{4}{3}x^2-\dfrac{16}{3}x-\dfrac{16}{3}-\dfrac{5}{4}x^2+\dfrac{5}{4}=\dfrac{3}{2}x^2-3x-x-4\)
\(\Leftrightarrow x^2\cdot\dfrac{-25}{12}-\dfrac{25}{3}x-\dfrac{103}{12}-\dfrac{3}{2}x^2+4x+4=0\)
\(\Leftrightarrow\dfrac{-43x^2}{12x}-\dfrac{13x}{3}-\dfrac{55}{12}=0\)
\(\Leftrightarrow43x^2+52x+55=0\)
\(\text{Δ}=52^2-4\cdot43\cdot55=-6756< 0\)
Do đó: Phương trình vô nghiệm
c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2+2x+2}{x+1}\right)^2=\left(x+1\right)^2+\left(x+1+\dfrac{1}{x+1}\right)^2=2\left(x+1\right)^2+\dfrac{1}{\left(x+1\right)^2}+2\ge_{AM-GM}2\sqrt{2}+2\).
Đẳng thức xảy ra khi \(2\left(x+1\right)^2=\dfrac{1}{\left(x+1\right)^2}\Leftrightarrow x=\pm\sqrt{\dfrac{1}{2}}-1\).
b) \(g\left(x\right)=\dfrac{\left(x+2\right)\left(x+3\right)}{x}=\dfrac{x^2+5x+6}{x}=\left(x+\dfrac{6}{x}\right)+5\ge_{AM-GM}2\sqrt{6}+5\).
Đẳng thức xảy ra khi x = \(\sqrt{6}\).
(x-1)^3-(x+3)(x^2-3x+9)+3(x^2-4)=2
=>x^3-3x^2+3x-1-x^3-27+3x^2-12=2
=>3x-40=2
=>x=42/3=14
a: \(y=\left(5x-10\right)^4\)
=>\(y'=4\cdot\left(5x-10\right)'\cdot\left(5x-10\right)^3\)
\(=4\cdot5\cdot\left(5x-10\right)^3=20\left(5x-10\right)^3\)
Đặt y'>0
=>\(20\left(5x-10\right)^3>0\)
=>\(\left(5x-10\right)^3>0\)
=>5x-10>0
=>x>2
Đặt y'<0
=>\(20\left(5x-10\right)^3< 0\)
=>\(\left(5x-10\right)^3< 0\)
=>5x-10<0
=>x<2
Vậy: hàm số đồng biến trên \(\left(2;+\infty\right)\)
Hàm số nghịch biến trên \(\left(-\infty;2\right)\)
c: \(y=\left(x^3-1\right)^3\)
=>\(y'=3\left(x^3-1\right)'\cdot\left(x^3-1\right)^2\)
\(=9x^2\left(x^3-1\right)^2>=0\forall x\)
=>Hàm số luôn đồng biến trên R
d: \(y=\left(x^2-1\right)\left(x+2\right)\)
=>\(y'=\left(x^2-1\right)'\left(x+2\right)+\left(x^2-1\right)\left(x+2\right)'\)
\(=2x\left(x+2\right)+x^2-1\)
\(=2x^2+4x+x^2-1=3x^2+4x-1\)
Đặt y'>0
=>\(3x^2+4x-1>0\)
=>\(\left[{}\begin{matrix}x< \dfrac{-2-\sqrt{7}}{3}\\x>\dfrac{-2+\sqrt{7}}{3}\end{matrix}\right.\)
Đặt y'<0
=>\(3x^2+4x-1< 0\)
=>\(\dfrac{-2-\sqrt{7}}{3}< x< \dfrac{-2+\sqrt{7}}{3}\)
Vậy: Hàm số đồng biến trên các khoảng \(\left(-\infty;\dfrac{-2-\sqrt{7}}{3}\right);\left(\dfrac{-2+\sqrt{7}}{3};+\infty\right)\)
Hàm số nghịch biến trên khoảng \(\left(\dfrac{-2-\sqrt{7}}{3};\dfrac{-2+\sqrt{7}}{3}\right)\)
b: \(y=\left(-x-1\right)\left(x+2\right)^4\)
=>\(y'=\left(-x-1\right)'\left(x+2\right)^4+\left(-x-1\right)\left[\left(x+2\right)^4\right]'\)
\(=-\left(x+2\right)^4+\left(-x-1\right)\cdot4\left(x+2\right)'\left(x+2\right)^3\)
\(=-\left(x+2\right)^4+4\left(x+2\right)^3\cdot\left(-x-1\right)\)
\(=-\left(x+2\right)^3\left[\left(x+2\right)+4\left(x+1\right)\right]\)
\(=-\left(x+2\right)^2\cdot\left(x+2\right)\left(5x+6\right)\)
Đặt y'<0
=>\(-\left(x+2\right)^2\left(x+2\right)\left(5x+6\right)< 0\)
=>(x+2)(5x+6)>0
TH1: \(\left\{{}\begin{matrix}x+2>0\\5x+6>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-2\\x>-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x>-\dfrac{6}{5}\)
TH2: \(\left\{{}\begin{matrix}x+2< 0\\5x+6< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -2\\x< -\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x< -2\)
Đặt y'>0
=>(x+2)(5x+6)<0
TH1: \(\left\{{}\begin{matrix}x+2>0\\5x+6< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-2\\x< -\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow-2< x< -\dfrac{6}{5}\)
TH2: \(\left\{{}\begin{matrix}x+2< 0\\5x+6>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -2\\x>-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy: HSĐB trên các khoảng \(\left(-\infty;-2\right);\left(-\dfrac{6}{5};+\infty\right)\)
HSNB trên khoảng \(\left(-2;-\dfrac{6}{5}\right)\)
\(2\left(x-1\right)+3\left(x-2\right)=x-4\)
\(\Rightarrow2x-2+3x-6=x-4\)
\(\Rightarrow2x+3x-x=-4+2+6\)
\(\Rightarrow4x=4\Rightarrow x=1\)
2(x-1)+3(x-2)=x-4
2.x-2.1+3.x-3.2=x-4
2x-2+3x-6=x-4
2x+3x-x=-4+2+6
4x=4
x=4:4
x=1
vậy x=1