x10 + 20x - 30x = 0
Những câu hỏi liên quan
rut gon (4x^4-20x^3+13x^2+30x+9)/(4x^2-1)^2
Giải PT: \(\sqrt{3x^2-30x+100}+\sqrt{8x^2-80x+216}=-2x^2+20x-41\)
tìm x biết:
1/2x +1/6x+1/12x+1/20x+1/30x+1/42x=36
=1/x*(1/2+1/6+1/12+1/20+1/30+1/42).
Ta có:
1/2+1/6+1/12+1/20+1/30+1/42.
=1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7.
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7.
=1-1/7.
=6/7.
=>1/x*6/7=36.
=>1/x=36:6/7=42.
=>x=1/42.
Vậy x=1/42.
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Bài 2: Tìm x
a) x³ - 16x =0
b) x^4 - 2x³ + 10x² - 20x = 0
c) ( 2x - 1)² = ( x + 3 )²
d) x²( x - 2 ) - 2x² + 8x - 8 = 0
e) x^4 - 30x² + 31x - 30 = 0
Bài 2 :
a ) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)
Vậy..........
b ) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\\x^2+10=0\left(loại\right)\end{matrix}\right.\)
Vậy .......................
c ) \(\left(2x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy.............
d ) \(x^2\left(x-2\right)-2x^2+8x-8=0\)
\(\Leftrightarrow x^3-2x^2-2x^2+8x-8=0\)
\(\Leftrightarrow x^3-4x^2+8x-8=0\)
\(\Leftrightarrow\) \(\left(x-2\right)^3=0\)
\(\Rightarrow x=2\)
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Bài 2 :
a ) x3−16x=0x3−16x=0
⇔x(x2−16)=0⇔x(x2−16)=0
⇔[x=0x2−16=0⇒[x=0x=±4⇔[x=0x2−16=0⇒[x=0x=±4
Vậy..........
b ) x4−2x3+10x2−20x=0x4−2x3+10x2−20x=0
⇔x3(x−2)+10x(x−2)=0⇔x3(x−2)+10x(x−2)=0
⇔(x−2)(x3+10x)=0⇔(x−2)(x3+10x)=0
⇔x(x−2)(x2+10)=0⇔x(x−2)(x2+10)=0
⇔⎡⎢⎣x=0x−2=0⇒x=2x2+10=0(loại)⇔[x=0x−2=0⇒x=2x2+10=0(loại)
Vậy .......................
c ) (2x−1)2=(x+3)2(2x−1)2=(x+3)2
⇔(2x−1)2−(x+3)2=0⇔(2x−1)2−(x+3)2=0
⇔(2x−1−x−3)(2x−1+x+3)=0⇔(2x−1−x−3)(2x−1+x+3)=0
⇔(x−4)(3x+2)=0⇔(x−4)(3x+2)=0
⇔[x−4=03x+2=0⇒⎡⎣x=4x=−23⇔[x−4=03x+2=0⇒[x=4x=−23
Vậy.............
d ) x2(x−2)−2x2+8x−8=0x2(x−2)−2x2+8x−8=0
⇔x3−2x2−2x2+8x−8=0⇔x3−2x2−2x2+8x−8=0
⇔x3−4x2+8x−8=0⇔x3−4x2+8x−8=0
⇔⇔ (x−2)3=0(x−2)3=0
⇒x=2
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4x^4-20x^3+13x^2+30X+9chia cho
(4x^2-1)^2
rut gon ho tui
Tìm GTNN
B= \(\sqrt{25x^2-20x+4}+\sqrt{25x^2-30x+9}\)
\(B=\left|5x-2\right|+\left|5x-3\right|\)
\(=\left|5x-2\right|+\left|3-5x\right|\)
=>B>=|5x-2+3-5x|=1
Dấu = xảy ra khi (5x-2)(5x-3)<=0
=>2/5<=x<=3/5
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tìm x biết :
1) x(x-5)-4x+20=0
2) x(x+6)-7x-42=0
3) x4-2x3+10x2-20x=0
4) x2+3x-18=0
5) 8x2+3x+7=0
6) x3-11x2+30x=0
Cho �3�163x16y. Tìm �x; �y biết �+�190x+y190.
�30x30; �160y160.
�15x15; �175y175.
�20x20; �170y170.
�10x10; �180y180.
Đọc tiếp
Cho . Tìm ; biết .
; . ; . ; . ; .
Lời giải:
$3x=16y\Rightarrow \frac{x}{16}=\frac{y}{3}$
Áp dụng TCDTSBN:
$\frac{x}{16}=\frac{y}{3}=\frac{x+y}{16+3}=\frac{190}{19}=10$
$\Rightarrow x=10.16=160; y=3.10=30$
Đáp án A.
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1.Tìm x:
a,x mũ 3 - 16 = 0
b,x mũ 4 - 2x mũ 3 + 10x mũ 2 - 20x = 0
c,(2x - 3)mũ 2 = (x + 5)mũ 2
d,x mũ 2(x - 1) - 4x mũ 2 + 8x -4 = 0
e,x mũ 3 - 11x mũ 2 + 30x = 0
P/s:Giúp mk vs chiều mk phải nộp rồi
b \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
hay \(x\in\left\{0;2\right\}\)
c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
=>(x-8)(3x+2)=0
=>x=8 hoặc x=-2/3
d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
=>x=2 hoặc x=1
e: \(\Leftrightarrow x\left(x^2-11x+30\right)=0\)
=>x(x-5)(x-6)=0
hay \(x\in\left\{0;5;6\right\}\)
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