không có đáp án nào chính xác

mình nghĩ thế thôi

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 \(\Leftrightarrow x=\dfrac{3}{4}\)

c) \(C=4x^2+40x-2\)

\(C=\left(2x\right)^2+2\cdot2x\cdot10+10^2-102\)

\(C=\left(2x+10\right)^2-102\ge-102\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-5\)

\(\Leftrightarrow\left(x^2-12x-6\right)\left(x^2-4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-12x-6=0\\x^2-4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-6\right)^2=42\\\left(x-2\right)^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6\in\left\{\sqrt{42};-\sqrt{42}\right\}\\x-2\in\left\{\sqrt{2};-\sqrt{2}\right\}\end{matrix}\right.\Leftrightarrow x\in\left\{\sqrt{42}+6;-\sqrt{42}+6;\sqrt{2}+2;2-\sqrt{2}\right\}\)

Đây đích thực có phải là lớp 1 ko bn?

Lên lớp trên mà gửi đi gửi ở lớp 1 làm gì vậy bạn .

\(20x^2+24x+18=500\)    

\(20x^2+24x-482=0\) 

\(10x^2+12x-241=0\) 

\(\orbr{\begin{cases}x=\frac{-6+\sqrt{2446}}{10}\\x=\frac{-6-\sqrt{2446}}{10}\end{cases}}\)

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 ;-;

Ta có: \(20x^2+24x+18=500\)

    \(\Leftrightarrow20x^2+24x-482=0\)

    \(\Leftrightarrow100x^2+120x-2410=0\)

    \(\Leftrightarrow\left(100x^2+120x+36\right)-2446=0\)

    \(\Leftrightarrow\left(10x+6\right)^2=2446\)

    \(\Leftrightarrow10x+6=\pm\sqrt{2446}\)

 + \(10x+6=\sqrt{2446}\)\(\Leftrightarrow x=\frac{\sqrt{2446}-6}{10}\approx4,345705208\)

 + \(10x+6=-\sqrt{2446}\)\(\Leftrightarrow x=\frac{-\sqrt{2446}-6}{10}\approx-5,545705208\)

Vậy \(S=\left\{4,345705208;-545705208\right\}\)

\(x+20\times x+10\times x+40\times x+30=205000\)

\(x\times\left(1+20+10+40\right)+30=205000\)

\(x\times71+30=205000\)

\(x\times71=205000-30\)

\(x\times71=204970\)

\(x=204970:71\)

\(x=\frac{204970}{71}\)

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

\(\left[{}\begin{matrix}x=\dfrac{-5+\sqrt[]{21}}{2}\\x=\dfrac{-5-\sqrt[]{21}}{2}\end{matrix}\right.\)