Gọi số cần tìm có dạng là \(X=\overline{ab}\)

Khi viết thêm chữ số 5 vào bên phải của số đó thì số mới lớn hơn số cần tìm là 689 đơn vị nên ta có:

\(\overline{ab5}-\overline{ab}=689\)

=>\(10\cdot\overline{ab}+5-\overline{ab}=689\)

=>\(9\cdot\overline{ab}=684\)

=>\(9\cdot X=684\)

=>\(X=\dfrac{684}{9}=76\)

Vậy: Số cần tìm là 76

\(a.\left(x+1\right)^2-\left(x-1\right)^2-3\left(x+1\right)\left(x-1\right)\\ =\left(x^2+2x+1\right)-\left(x^2-2x+1\right)-3\left(x^2-1\right)\\ =x^2+2x+1-x^2+2x-1-3x^2+3\\ =4x-3x^2+3\\b.5\left(x-2\right)\left(x+2\right)-\dfrac{1}{2}\left(6-8x\right)^2+17\\ =5\left(x^2-4\right)-\dfrac{1}{2}\left(36-96x+64x^2\right)+17\\ =5x^2-20-18+48x-32x^2\\ =48x-27x^2-38\)

\(2,8\cdot\dfrac{-6}{13}-7,2-2,8\cdot\dfrac{7}{13}\\ =\left(2,8\cdot\dfrac{-6}{13}-2,8\cdot\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\left(\dfrac{-6}{13}-\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\dfrac{-13}{13}-7,2\\=-2,8-7,2\\ =-10\)

Giúp mình với ạ

\(2,8\cdot\dfrac{-6}{13}-7,2-2,8\cdot\dfrac{7}{13}\\ =\left(2,8\cdot\dfrac{-6}{13}-2,8\cdot\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\left(\dfrac{-6}{13}-\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\dfrac{-13}{13}-7,2\\ =-2,8-7,2\\ =-10\)

\(a.\left(x+y+4\right)\left(x+y-4\right)\\ =\left[\left(x+y\right)+4\right]\left[\left(x+y\right)-4\right]\\ =\left(x+y\right)^2-4^2\\ b.\left(x-y+6\right)\left(x+y-6\right)\\ =\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\\ =x^2-\left(y-6\right)^2\\ c.\left(y+2z-3\right)\left(y-2z-3\right)\\ =\left[\left(y-3\right)+2z\right]\left[\left(y-3\right)-2z\right]\\ =\left(y-3\right)^2-\left(2z\right)^2\\ d.\left(x+2y+3z\right)\left(2y+3z-x\right)\\ =\left[\left(2y+3z\right)+x\right]\left[\left(2y+3z\right)-x\right]\\ =\left(2y+3z\right)^2-x^2\)

\(\left(x+1\right)^2-\left(x-1\right)^2\\ =\left[\left(x+1\right)-\left(x-1\right)\right]\left[\left(x+1\right)+\left(x-1\right)\right]\\ =\left(x+1-x+1\right)\left(x+1+x-1\right)\\ =2\cdot2x\\ =4x\)

\(x:0,25+x\times11=24\)

\(x\times4+x\times11=24\)

\(x\times\left(4+11\right)=24\)

\(x\times15=24\)

\(x=24:15\)

\(x=1,6\)

M = {\(x\in\) Q/ \(x\) = \(\dfrac{k}{k+1}\);  6 ≥ k \(\in\) N*}

K = {\(x\) \(\in\) Q/ \(x\) = \(\dfrac{k}{k+3}\); 6 ≥ k \(\in\) N*}

\(a.5\cdot3^x=5\cdot3^4\\ =>3^x=\dfrac{5\cdot3^4}{5}=3^4\\ =>x=4\\ b.7\cdot4^x=7\cdot4^3\\ =>4^x=\dfrac{7\cdot4^3}{7}=4^3\\ =>x=3\\ c.\dfrac{3}{5}\cdot4^x=7\cdot4^3\\ =>4^x=\dfrac{7\cdot4^3}{\dfrac{3}{5}}=\dfrac{35}{3}\cdot4^3\\ =>\dfrac{4^x}{4^3}=\dfrac{35}{3}\\ =>4^{x-3}=\dfrac{35}{3}\\ =>x-3=log_4\dfrac{35}{3}\\ =>x=log_4\dfrac{35}{3}+3\\ d.\dfrac{3}{2}\cdot5^x=\dfrac{3}{2}\cdot5^{12}\\ =>5^x=\dfrac{5^{12}\cdot\dfrac{3}{2}}{\dfrac{3}{2}}=5^{12}\\ =>x=12\) 

e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)

=>\(9\cdot5^x=9\cdot5^6\)

=>\(5^x=5^6\)

=>x=6

f: \(5\cdot3^x=7\cdot3^5-2\cdot3^5\)

=>\(5\cdot3^x=5\cdot3^5\)

=>\(3^x=3^5\)

=>x=5

g: \(5\cdot3^{x+6}=2\cdot3^5+3\cdot3^5\)

=>\(5\cdot3^{x+6}=5\cdot3^5\)

=>\(3^{x+6}=3^5\)

=>x+6=5

=>x=-1

`(x+1)^2 + (x-1)^2`

`= x^2 + 2x + 1 + x^2 - 2x + 1`

`= 2x^2 + 2`

`= 2(x^2 +1)`

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Áp dụng hằng đẳng thức: 

\(\left(a\pm b\right)^2=a^2\pm2ab+b^2\)