mik chẳng hỉu gì cả

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bằng 1024

Mik bấm bừa là đúng :v

DO khong co dieu kien cua x nen ban hay lay x la mot so tu nhien bat ki

giả sử lấy x=1 thì ta có thể dễ dàng tính được tổng bằng 4^5=1024

a) \(7x^2-28=0\Leftrightarrow7\left(x^2-4\right)=0\Leftrightarrow x^2-4=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) vậy \(x=2;x=-2\)

b) \(\left(2x+1\right)+x\left(2x+1\right)=0\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\2x=-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=\dfrac{-1}{2}\end{matrix}\right.\) vậy \(x=-1;x=\dfrac{-1}{2}\)

c) \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-5=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\) vậy \(x=0;x=5;x=-5\)

d) \(9\left(3x-2\right)=x\left(2-3x\right)\Leftrightarrow9\left(3x-2\right)=-x\left(3x-2\right)\)

\(\Leftrightarrow9\left(3x-2\right)+x\left(3x-2\right)=0\Leftrightarrow\left(9+x\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}9+x=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\3x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x=-9;x=\dfrac{2}{3}\)

e) \(5x\left(x-3\right)-2x+6=0\Leftrightarrow5x\left(x-3\right)-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(5x-2\right)\left(x-3\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}5x-2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\) vậy \(x=\dfrac{2}{5};x=3\)

a) (2x-5)(x-1)<0 nên âm

=> 2x-5>0 ; x-1<0 hoặc 2x-5<0 ; x-1>0

=> 2x>5 ; x<1 hoặc 2x<5 ; x>1

=> x>=2 ; x<1 hoặc x<=2 ; x>1

chúc bnaj học tốt!!!

a) 

Để \(\left(3x-1\right).\left(-\frac{1}{2}x+5\right)=0\)=> 3x-1=0 hoặc \(-\frac{1}{2}x+5=0\)

=> x= \(\frac{1}{3}\) hoăc \(x=10\)

b)

\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=5\) => \(\frac{1}{3}:\left(2x-1\right)=5-\frac{1}{4}=\frac{19}{4}=>2x-1=\frac{1}{3}:\frac{19}{4}=\frac{4}{57}=>x=\frac{61}{114}\)

c) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0=>\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)\(=>2x+\frac{3}{5}\in\left\{\pm\frac{3}{5}\right\}=>2x\in\left\{0;\frac{-6}{5}\right\}=>x\in\left\{0;\frac{-3}{5}\right\}\)

d) Xem lại đề

a) để (3x-1).(\(-\dfrac{1}{2}x+5\))=0

=> 3x-1 hoặc \(-\dfrac{1}{2}x+5\) =0

TH1 : 3x-1=0

3x = 0+1=1

x = 1:3 = \(\dfrac{1}{3}\)

TH2 : \(-\dfrac{1}{2}x+5\)= 0

\(-\dfrac{1}{2}x\)= 0 -5 = -5

x= -5 : \(-\dfrac{1}{2}\)

x= 10

d. (3x-\(\frac{1}{2}\))\(^3\) = -\(\frac{1}{9}:3=-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)

=> \(3x-\frac{1}{2}=\frac{-1}{3}\)

=> 3x = \(\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

=> x= \(\frac{1}{18}\)

a: \(x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

=>x=1 hoặc x=3

b: \(x^2+x-12=0\)

=>(x+4)(x-3)=0

=>x=3 hoặc x=-4

c: \(3x^2+2x-5=0\)

\(\Leftrightarrow3x^2+5x-3x-5=0\)

=>(3x+5)(x-1)=0

=>x=1 hoặc x=-5/3

d: \(x^4-2x^2-3=0\)

\(\Leftrightarrow x^4-3x^2+x^2-3=0\)

\(\Leftrightarrow x^2-3=0\)

hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)

a)-6

b)-6

c)0.3652593485...

ủng hộ neji đê mọi người ơi

a=0

b= -1/3

c=1

Ace LegonaĐặng Phương Nam

x=-3 hoặc x=20