\(x^2+6x-7=0\)

\(\Leftrightarrow x^2-x+7x-7=0\)

\(\Leftrightarrow x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+7=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=-7\end{cases}}\)

Ta có:
\(x^3+5x^2+3x-9=0\)

\(\Leftrightarrow x^3+3x^2+2x^2+6x-3x-9=0\)

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

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

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

\(\Leftrightarrow\left(x+3\right)^2\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+3=0\\x=1\end{cases}\orbr{\begin{cases}x=-3\\x=1\end{cases}}}}\)

Dạng kiểu này bạn dùng phương pháp nhẩm nghiệm 

\(x^3+5x^2+3x-9=0\)

\(\Leftrightarrow x^3+4x^2+x^2+3x-9=0\)

\(\Leftrightarrow\left(x^3+4x^2+3x\right)+\left(x^2-9\right)=0\)

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

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

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

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

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

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

\(\Leftrightarrow\left(x+3\right)^2\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}}\)

Vậy tập nghiệm của pt là S={-3;1}

_Học tốt_

=>1-0,5

=>0,5

\(1-\left(\frac{2}{5}+\frac{1}{10}\right)=1-\left(\frac{4}{10}+\frac{1}{10}\right)=1-\frac{1}{2}=\frac{1}{2}\)

1-(2/5+1/10)=1-1/2=1/2

nha bnk cho mk nkamỹ hoàng

C1 : 5/24 x 5/12 x 24

=    5/24 x ( 5/12 x 24 ) 

=   5/24  x   10 

=         50/24

=         25/12

C2 :            5/24 x 5/12 x 24

=             ( 5/24 x 24 ) x ( 5/12 ) 

=                    5 x (  5/12 )  

=                25/12

Chúc em học giỏi !!! 

Cách 1 : 5/24 x 5/12 x 24 = (5/24 x 24) x 5/12

                                        = 5 x 5/12 = 25/12

Cách 2 : 5/24 x 5/12 x 24 = 5/24 x (5/12 x 24)

                                         = 5/24 x 10 = 25/12

c1

5/24x5/12x24

=5/24x(5/12x24)

=5/24x10

=50/24

=25/12

c2

5/24x5/12x24

=(5/24x24)x5/12

=5x5/12=25/12

~~ học tốt nhé!!

=> x²  = 0+49

=>x²  = 49

=>x²=7²

=>x=7

vậy x=7

\(A=\left(x-5\right)^2+\left(x+1\right)^2+5=x^2-10x+25+x^2+2x+1+5.\)

\(=2x^2-8x+31=2\left(x^2-4x\right)+31=2\left(x^2-2.x.2+4\right)-8+31\)

\(=2\left(x-2\right)^2+23\)

Vì \(2\left(x-2\right)^2\ge0\forall x\)nên \(2\left(x-2\right)^2+23\ge23\forall x\)

Vậy \(MinA=23\Leftrightarrow2\left(x-2\right)^2=0\Leftrightarrow x=2\)

lỗi phép tính r nha e

a,\(x\in\left\{5;1,5;\dfrac{-4}{3}\right\}\)

\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)

\(4x^2-40x+100=18x+9\)

\(4x^2-58x+91=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)

\(c,x^3+3x^2-6x-8=0\)

\(\left(x+4\right)\left(x-2\right)\left(x+1\right)=0\)

\(Th1:x+4=0\Leftrightarrow x=-4\)

\(Th2:x-2=0\Leftrightarrow x=2\)

\(Th3:x+1=0\Leftrightarrow x=-1\)

\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)

\(< =>4x^2-40x+100=18x+9\)

\(< =>4x^2+58x+91=0\)

\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)