Gọi H=abc-cba

<=>H=100*a+10*b+c-100*c-10*b-c

<=>H=99*a-99*c

<=>H=99*(a-c) (1)

thay a-c=4 vào (1) 

(1)<=> H=99*4=396

Vậy abc-cba=396

396 giải ở câu trên

 1x 32 + 7 x 8 + 4x = 200x 

t ưu tiên cho phép nhân : 

32 + 56 = 196x <=> 88 = 196x <=> x = 88 / 196 <=> x = 22 / 49

Vậy.....

1y32+7y8+4y=200y

a)Ta có:7=7.1  ;   9=3.3   ;     11=11.1

=>BC(7;9;11)=1

b)Ta có:24=(2^3).3   ;    28=(2^2).7:        ;40=(2^3).5

=>BC(24;28;40)=2^2=4

Nhớ giữ lời hứa nhé

Mình nhầm.

a)7=7.1   ;   9=(3^2)  ;    11=11.1

BC(7;9;11)=7.(3^2).11=693

b)Ta có:24=(2^3).3   ;   28=(2^2).7  ;   40=(2^3).5

BC(24;28;40)=2^3.3.5.7=840

Xin lỗi nhé

a, 

ta có :  7 = 7 

            9 = \(^{3^2}\) 

             11 = 11 

BC ( 7 ,  9 , 11 )  = 1

b ,

ta có :   24 = \(2^3.3\)

             28 = \(2^2.7\) 

            40 = \(2^3.5\) 

BCNN  ( 24 ,28 ,40 ) = \(2^3.3.5.7\) =  840

BC  ( 24 ,28 ,40 ) = B ( 840 ) = { 0 ; 840 ; 1680 ;...}

`a,-2022.(x+8)=0`

`=>x+8=0:(-2022)`

`=>x+8=0`

`=>x=0-8`

`=>x=-8`

`b,(7-x)(x+3)=0`

`@ TH1`

`7-x=0`

`=>x=7-0`

`=>x=7`

`@ TH2`

`x+3=0`

`=>x=0-3`

`=>x=-3`

`c,2023x .(14-x)=0`

`@ TH1`

`2023x=0`

`=>x=0:2023`

`=>x=0`

`@ TH2`

`14-x=0`

`=>x=14-0`

`=>x=14`

`d,x^2-x=0`

`=>x(x-1)=0`

`@ TH1`

`x=0`

`@ TH2`

`x-1=0`

`=>x=0+1`

`=>x=1`

\(a,\left(-2022\right).\left(x+8\right)=0\\ \Rightarrow\left(x+8\right)=0:\left(-2022\right)\\ \Rightarrow x+8=0\\ \Rightarrow x=-8\\ b,\left(7-x\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right. \Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\\ c,2023x.\left(14-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2023x=0\\14-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\\ d,x^2-x=0\\ \Rightarrow x\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(\left(-2020\right).\left(x+8\right)=0\)

\(=>x+8=0=>x=-8\)

\(\left(7-x\right).\left(x+3\right)=0\)

\(=>\left[{}\begin{matrix}7-x=0\\x+3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)

\(2023x.\left(14-x\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\14-x=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=14\end{matrix}\right.\)

\(x^2-x=0\)

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

\(=>\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

1032 + X x 100 + 708 + X x 10 + 40 + X x 1 = 2000 + X

(1032 + 708 + 40) + (X x 100 + X x 10 + X x 1) = 2000 + X

1780 + X x ( 100 + 10 +1 ) = 2000 + X

1780 + X x 111 =2000 + x

1780 + X x 111 - X = 2000

1780 + X x 110 =2000

X x 110 = 2000 - 1780

X x 110 = 220

X = 220 :110

X = 2

1032+708+40+(x00+x0+x)=2000+x.

=>1780+xx0+x=2000+x.

=>xx0=2000-1780=220.

=>x=2.

Vậy x=2.

tk mk nha các bn.

-chúc ai tk mk học giỏi và may mắn-

32x+56x+4x=200x

x(32+56+4)=200x

x92=200x

=>x92-200x=0

x(92-200)=0

x.(-108)=0

x=0:(-108)

x=0

\(\left(\frac{3}{5}\right)^2=\frac{9}{25}\)

\(n=5.2=10\)

\(\text{chúc e hok tốt nhé}\)

bài nay fdeex mà p

\(\left(\frac{3}{5}\right)^n=\left(\frac{9}{25}\right)^5\)

=> \(\frac{3^n}{5^n}=\frac{9^5}{25^5}\)

=> \(\frac{3^n}{5^n}=\frac{\left(3^2\right)^5}{\left(5^2\right)^5}\)

=> \(\frac{3^n}{5^n}=\frac{3^{10}}{5^{10}}\)

=> n = 10

x=0

Không tồn tại nghiệm số thực nhé bạn (88+4x=200x sai nhé )