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a: Ta có: \(x^2\ge0\forall x\)
\(\left(y-\dfrac{1}{10}\right)^4\ge0\forall y\)
Do đó: \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(0;\dfrac{1}{10}\right)\)
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`@` `\text {Ans}`
`\downarrow`
`a)`
Thu gọn:
`P(x)=`\(5x^4 + 3x^2 - 3x^5 + 2x - x^2 - 4 +2x^5\)
`= (-3x^5 + 2x^5) + 5x^4 + (3x^2 - x^2) + 2x - 4`
`= -x^5 + 5x^4 + 2x^2 + 2x - 4`
`Q(x) =`\(x^5 - 4x^4 + 7x - 2 + x^2 - x^3 + 3x^4 - 2x^2\)
`= x^5 + (-4x^4 + 3x^4) - x^3 + (x^2 - 2x^2) + 7x - 2`
`= x^5 - x^4 - x^3 - x^2 + 7x - 2`
`@` Tổng:
`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)
`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`
`= (-x^5 + x^5) - x^3 + (5x^4 - x^4) + (2x^2 - x^2) + (2x + 7x) + (-4-2)`
`= 4x^4 - x^3 + x^2 + 9x - 6`
`@` Hiệu:
`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)
`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`
`= (-x^5 - x^5) + (5x^4 + x^4) + x^3 + (2x^2 + x^2) + (2x - 7x) + (-4+2)`
`= -2x^5 + 6x^4 + x^3 + 3x^2 - 5x - 2`
`b)`
`@` Thu gọn:
\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)
`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`
`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`
`= x^4 - 2x^3 - x^2 + 15x + 10`
\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)
`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`
`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`
`= x^4 + 3x^3 + 2x - 4`
`@` Tổng:
`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)
`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`
`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`
`= 2x^4 + x^3 - x^2 + 17x + 6`
`@` Hiệu:
`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)
`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`
`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`
`= -5x^3 - x^2 + 13x + 14`
`@` `\text {# Kaizuu lv u.}`
\(\left(x-3\right)\left(4-x\right)>0\)
\(\Rightarrow\)\(\hept{\begin{cases}x-3>0\\4-x>0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x>3\\x< 4\end{cases}}\) (vô lí)
hoặc \(\hept{\begin{cases}x-3< 0\\4-x< 0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x< 3\\x>4\end{cases}}\)(vô lí)
Vậy \(x=\Phi\)
Lời giải:
1.
$4x+9=0$
$4x=-9$
$x=\frac{-9}{4}$
2.
$-5x+6=0$
$-5x=-6$
$x=\frac{6}{5}$
3.
$x^2-1=0$
$x^2=1=1^2=(-1)^2$
$x=\pm 1$
4.
$x^2-9=0$
$x^2=9=3^2=(-3)^2$
$x=\pm 3$
5.
$x^2-x=0$
$x(x-1)=0$
$x=0$ hoặc $x-1=0$
$x=0$ hoặc $x=1$
6.
$x^2-2x=0$
$x(x-2)=0$
$x=0$ hoặc $x-2=0$
$x=0$ hoặc $x=2$
7.
$x^2-3x=0$
$x(x-3)=0$
$x=0$ hoặc $x-3=0$
$x=0$ hoặc $x=3$
8.
$3x^2-4x=0$
$x(3x-4)=0$
$x=0$ hoặc $3x-4=0$
$x=0$ hoặc $x=\frac{4}{3}$
b: 4x^2-20x+25=(x-3)^2
=>(2x-5)^2=(x-3)^2
=>(2x-5)^2-(x-3)^2=0
=>(2x-5-x+3)(2x-5+x-3)=0
=>(3x-8)(x-2)=0
=>x=8/3 hoặc x=2
c: x+x^2-x^3-x^4=0
=>x(x+1)-x^3(x+1)=0
=>(x+1)(x-x^3)=0
=>(x^3-x)(x+1)=0
=>x(x-1)(x+1)^2=0
=>\(x\in\left\{0;1;-1\right\}\)
d: 2x^3+3x^2+2x+3=0
=>x^2(2x+3)+(2x+3)=0
=>(2x+3)(x^2+1)=0
=>2x+3=0
=>x=-3/2
a: =>x^2(5x-7)-3(5x-7)=0
=>(5x-7)(x^2-3)=0
=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)
`#3107.101107`
`1/2x + 4/5 = 2x - 8/5`
`=> 1/2x - 2x = -4/5 - 8/5`
`=> -3/2x = -12/5`
`=> x = -12/5 \div (-3/2)`
`=> x = 8/5`
Vậy, `x = 8/5`
_____
`\sqrt{x} = 5`
`=> x = 5^2`
`=> x = 25`
Vậy, `x = 25`
___
`x^2 = 3`
`=> x^2 = (+-\sqrt{3})^2`
`=> x = +- \sqrt{3}`
Vậy, `x \in {-\sqrt{3}; \sqrt{3}}.`
Bài 1:
a) \(3x^2\left(2x^3-x+5\right)-6x^5-3x^3+10x^2\)
\(=6x^5-3x^3+10x^2-6x^5-3x^3+10x^2\)
\(=10x^2+10x^2\)
\(=20x^2\)
b) \(-2x\left(x^3-3x^2-x+11\right)-2x^4+3x^3+2x^2-22x\)
\(=-2x^4+6x^3+2x^2-22x-2x^4+3x^3+2x^2-22x\)
\(=-4x^4+9x^3+4x^2-44x\)
a ,( x2 -5 ) x ( x2 +9) x( -11-8x) =0
=> x2 -5 = 0 ; x2 + 9 = 0 hoặc -11-8 x =0 .
ko biết câu b)
xin lỗi !