Tìm x :
16x + 20x + 24x + .............. + 40x + 44x = 4800
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Lời giải:
$16x^3y^2-24x^2y^3+20x^4=16x^2(xy^2-\frac{3}{2}y^3+\frac{5}{4}x^2)$
$\Rightarrow 16x^3y^2-24x^2y^3+20x^4\vdots 16x^2$
Đáp án C.
a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)
=>(-2x+12)(4x+12)=0
=>x=-3 hoặc x=6
b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)
=>\(x\simeq0.93\)
d: =>-4x+28+11x=-x+3x+15
=>7x+28=2x+15
=>5x=-13
=>x=-13/5
e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)
=>-9x=-3x+5
=>-6x=5
=>x=-5/6
a) \(A=9x^2-30x+30\)
\(A=\left(3x\right)^2-2\cdot3x\cdot5+5^2+5\)
\(A=\left(3x-5\right)^2+5\ge5\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{3}\)
b) \(B=16x^2-24x-3\)
\(B=\left(4x\right)^2-2\cdot4x\cdot3+3^2-13\)
\(B=\left(4x-3\right)^2-13\ge-13\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{3}{4}\)
\(\Leftrightarrow\left(x^2-12x-6\right)\left(x^2-4x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-6\right)^2=42\\\left(x-2\right)^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6\in\left\{\sqrt{42};-\sqrt{42}\right\}\\x-2\in\left\{\sqrt{2};-\sqrt{2}\right\}\end{matrix}\right.\Leftrightarrow x\in\left\{\sqrt{42}+6;-\sqrt{42}+6;\sqrt{2}+2;2-\sqrt{2}\right\}\)
\(20x^2+24x+18=500\)
\(20x^2+24x-482=0\)
\(10x^2+12x-241=0\)
\(\orbr{\begin{cases}x=\frac{-6+\sqrt{2446}}{10}\\x=\frac{-6-\sqrt{2446}}{10}\end{cases}}\)
20x2 + 24x + 18 = 500
<=> 20x2 + 24x + 18 - 500 = 0
<=> 20x2 + 24x - 482 = 0
<=> 2( 10x2 + 12x - 241 ) = 0
<=> 10x2 + 12x - 241 = 0 (*)
\(\Delta'=b'^2-ac=\left(\frac{b}{2}\right)^2-ac=6^2-10\cdot\left(-241\right)=36+2410=2446\)
\(\Delta'>0\)nên (*) có hai nghiệm phân biệt :
\(\hept{\begin{cases}x_1=\frac{-b'+\sqrt{\Delta'}}{a}=\frac{-6+\sqrt{2446}}{10}\\x_2=\frac{-b'-\sqrt{\Delta'}}{a}=\frac{-6-\sqrt{2446}}{10}\end{cases}}\)
Lớp 8 sao nghiệm xấu thế nhỉ ;-;
\(x+20\times x+10\times x+40\times x+30=205000\)
\(x\times\left(1+20+10+40\right)+30=205000\)
\(x\times71+30=205000\)
\(x\times71=205000-30\)
\(x\times71=204970\)
\(x=204970:71\)
\(x=\frac{204970}{71}\)
a) \(4x^2+16x+3=0\)
\(\Delta'=84-12=72\Rightarrow\sqrt[]{\Delta'}=6\sqrt[]{2}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+6\sqrt[]{2}}{4}\\x=\dfrac{-8-6\sqrt[]{2}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\left(4-3\sqrt[]{2}\right)}{4}\\x=\dfrac{-2\left(4+3\sqrt[]{2}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(4-3\sqrt[]{2}\right)}{2}\\x=\dfrac{-\left(4+3\sqrt[]{2}\right)}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt[]{2}-4}{2}\\x=\dfrac{-3\sqrt[]{2}-4}{2}\end{matrix}\right.\)
b) \(7x^2+16x+2=1+3x^2\)
\(4x^2+16x+1=0\)
\(\Delta'=84-4=80\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{5}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+4\sqrt[]{5}}{4}\\x=\dfrac{-8-4\sqrt[]{5}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\left(2-\sqrt[]{5}\right)}{4}\\x=\dfrac{-4\left(2+\sqrt[]{5}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\left(2-\sqrt[]{5}\right)\\x=-\left(2+\sqrt[]{5}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt[]{5}\\x=-2-\sqrt[]{5}\end{matrix}\right.\)
c) \(4x^2+20x+4=0\)
\(\Leftrightarrow4\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow x^2+5x+1=0\)
\(\Delta=25-4=21\Rightarrow\sqrt[]{\Delta}=\sqrt[]{21}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-5+\sqrt[]{21}}{2}\\x=\dfrac{-5-\sqrt[]{21}}{2}\end{matrix}\right.\)
16x+20x+24x+................+40x+44x=4800
=(16+20+24+.....+40+44)x=4800
=240x=4800
x=20
16x + 20x + 24x + .............. + 40x + 44x = 4800
(16 + 20 + 24 + ..................+ 40 + 44). x = 4800
\(\frac{\left(16+44\right).8}{2}\). x = 4800
240 . x = 4800
x = 4800 : 240
x = 20
đáp số: 20