Bài 2:

Gọi số ban đầu là \(\overline{ab}\)

Theo đề, ta có: 5a+2b=29 và 10b+a-10a-b=36

=>5a+2b=29 và -9a+9b=36

=>a=3 và b=7

\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)

\(10,\left(x+3\right)^2-x^2=45\)

\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)

Vậy \(S=\left\{6\right\}\)

\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)

\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)

b) 3x - 6 - (8x + 4) - (10x + 15) = 50

=> 3x - 6 - 8x - 4 - 10x - 15  = 50

=> (3x - 8x - 10x)  =  6+ 4 + 15 + 50

=> -15x = 75 => x = 75 : (-15) = -5

c) => 2x - 3 = 2 - x hoặc 2x - 3 = - (2 - x) (Vì 2 số  có giá trị tuyệt đối bằng nhau thì chings bằng nhau hoặc đối nhau)

+) nếu 2x - 3 = 2 - x => 2x+ x = 2 + 3 => 3x = 5 => x = 5/3

+) nếu 2x - 3 = -(2 - x) => 2x - 3 = -2 + x => 2x - x = -2 + 3 => x = 1

Vậy x = 5/3 hoặc x = 1

a) (n-1)ⁿ⁺¹¹-(n-1)ⁿ=0

(n-1)ⁿ(n-1)¹¹-(n-1)ⁿ=0

(n-1)ⁿ[(n-1)¹¹-1]=0

(n-1)ⁿ=0 hoặc (n-1)¹¹-1=0

n-1=0   hoặc  (n-1)¹¹   =1

n=1      hoặc  n-1         =1

n=1      hoặc   n          =2

a) 3x + 2(5 - x) = -11

=> 3x + 10 - 2x = -11

=> 3x - 2x + 10 = -11

=> x = -21

b) 3x² - 3x(-2 + x) = 36

=> 3x² - 3x.(-2) - 3x.x = 36

=> 3x² + 6x - 3x² = 36

=> 6x = 36

=> x = 6

c) x(5 - 2x) + 2x(x - 1) = 15

=> x.5 + x.(-2x) + 2x.x + 2x.(-1) = 15

=> 5x - 2x² + 2x² - 2x = 15

=> 3x = 15

=> x = 5

Trả lời:

a,\(3x+2.\left(5-x\right)=-11\)

\(\Leftrightarrow3x+10-2x=-11\)

\(\Leftrightarrow x+10=-11\)

\(\Leftrightarrow x=-21\)

Vậy \(x=-21\)

b,\(3x^2-3x.\left(-2+x\right)=36\)

\(\Leftrightarrow3x^2+6x-3x^2=36\)

\(\Leftrightarrow6x=36\)

\(\Leftrightarrow x=6\)

Vậy \(x=6\)

c, \(x.\left(5-2x\right)+2x.\left(x-1\right)=15\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=15\)

\(\Leftrightarrow3x=15\)'

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

a) 3x + 2( 5 - x ) = -11

<=> 3x + 10 - 2x = -11

<=> x + 10 = -11

<=> x = -21

b) 3x² - 3x( -2 + x ) = 36

<=> 3x² + 6x - 3x² = 36

<=> 6x = 36

<=> x = 6

c) x( 5 - 2x ) + 2x( x - 1 ) = 1

<=> 5x - 2x² + 2x² - 2x = 15

<=> 3x = 15

<=> x = 3

a: =>11(x-3)=6(x-5)

=>11x-33=6x-30

=>5x=3

=>x=3/5

b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3

=>-9/4x+11/12=8/3

=>-9/4x=32/12-11/12=21/12=7/4

=>x=-7/9

c: =>1/2x-1/3-2/3x-1=x

=>-1/6x-4/3=x

=>-7/6x=4/3

=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7

d: =>1-2x-3x+1=7/2

=>-5x=3/2

=>x=-3/10

\(49\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)

\(4x^2+8x+4+4x^2-4x+1-8x^2+8-11=0\)

\(4x+2=0\)

\(4x=2\)

\(x=-\frac{1}{2}\)

<=>4(x²+2x+1)+4x²-4x+1-8x²+8-11=0

<=>4x²+8x+4+4x²-4x+1-8x²+8-11=0

<=>4x+2=0

<=>2(2x+1)=0

<=>2x+1=0

<=>x=-1/2

XL mik nhầm

\(=\frac{1}{2}\)

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