\(5^{x+1}-5^x=100\cdot25^{29}\)

\(\Rightarrow5^x\left(5-1\right)=100\cdot\left(5^2\right)^{29}\)

\(\Rightarrow5^x\cdot4=100\cdot5^{58}\)

\(\Rightarrow5^x=\dfrac{100\cdot5^{58}}{4}\)

\(\Rightarrow5^x=25\cdot5^{58}\)

\(\Rightarrow5^x=5^{60}\)

\(\Rightarrow x=60\)

\(5^{x+1}-5x=100.25^{29}\)

\(5.5^x-5^x=4.25.25^{29}\)

\(5^x.\left(5-1\right)4.25^{30}\)

\(4.5^x-4.\left(5^2\right)^{30}\)

\(5x=5^{60}\)

\(x=60\)

\(5^{x+1}-5^x=100.25^{29}\)

\(5^x.5-5^x=\left(5^2\right)^{29}.100\)

\(5^x.\left(5-1\right)=5^{2.29}.100\)

\(5^x.4=5^{58}.100\)

\(5^x=5^{58}.100:4\)

\(5^x=5^{58}.25\)

\(5^x=5^{58}.5^2\)

\(5^x=5^{60}\)

\(\Rightarrow x=60\)

=x*3+(20+2+1)=5x

=x*3+23=5x

=x*3+23=50+x

=x*3-x+23=50

=x*2+23=50

=x*2=50-23

=x*2=27

x=27/2

x=13.5

chọn mình nha bn

Ta có 5x + 7 = 5( x + 1 ) + 2

Để 5x + 7 chia hết cho x + 1

=> 5( x + 1 ) + 2 chia hết cho x + 1 

=>2 chia hết cho x + 1

=> x + 1 c Ư(2) = { 1 ; 2 ; -1 ; -2 }

=> x c { 0 ; 1 ; -2 ; -3 }

Vậy...

\(\frac{5x+7}{x+1}=\frac{5\left(x+1\right)+2}{x+1}=5+\frac{2}{x+1}\)

Để 5x+7\(⋮\)x+1 thì x+1\(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Xét bảng ( tự xét )

KL

\(5x+7⋮x+1\)

\(5\left(x+1\right)+6⋮x+1\)

Vì \(5\left(x+1\right)⋮x+1\)

\(6⋮x+1\)

\(\Rightarrow x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Ta lập bảng 

x = 5 hoặc x = -5

4/5x=3/5:-1/4

=>4x/5=-12/5

=>20x=-60

=>x=-60:20=-3

\(5x-1⋮2x+2\Leftrightarrow10x-2⋮2x+2\)

\(\Leftrightarrow5\left(2x+2\right)-12⋮2x+2\Leftrightarrow-12⋮2x+2\)

\(\Rightarrow2x+2\inƯ\left(-12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

tự làm nốt nhé ! 

\(\dfrac{5x-10}{77x^2+1}=0\)

Mà: \(77x^2+1\ge1>0\forall x\) 

\(\Rightarrow5x-10=0\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=\dfrac{10}{5}\)

\(\Rightarrow x=2\)

\(\left(5x-10\right):\left(77x^2+1\right)=0\)

\(TH1:5x-10=0\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=10:5\)

\(\Rightarrow x=2\)

\(TH2:77x^2+1=0\)

\(\Rightarrow77x^2=-1\) \(\left(vô.lý\right)\)

\(\Rightarrow\left(-1\ne0\right)\) 

Vậy nghiệm của đa thức \(\left(5x-10\right):\left(77x^2+1\right)=0\) là: \(x=2\)

\(\left(3x-1\right)\left(4x-1\right)\left(5x-1\right)\left(6x-1\right)=\left[\left(3x-1\right)\left(6x-1\right)\right]\left[\left(4x-1\right)\left(5x-1\right)\right]\)

\(=\left[18x^2-9x+1\right]\left[20x^2-9x+1\right]\)=120.

Do x thuộc Z => 2 cái trong ngoặc thuộc Ư(120) 

Đồng thời chúng đồng âm và đồng dương, Tất nhiên là cùng chẵn