=2768 dư 1

124565 : 45 = 24913/9 = 2768,1111111

2768(R=5)

Ta có:-(3-0,2.x)-80%=7,5

-3+0,2x-0,8=7,5

0,2x=7,5+3+0,8

X=11,3:0,2

X=56,5

Vậy x=56,5

Ta có:-(3-0,2.x)-80%=7,5

-3+0,2x-0,8=7,5

0,2x=7,5+3+0,8

X=11,3:0,2

X=56,5

Vậy x=56,5

3^a + 9^b  = 138

= 3^a + (3 x 3)^b = 138

= 3^a + 3^{2 +b} = 138

=3^{a + b + 2} = 138

còn lại cậu tự làm nhé

đi hỏi nhà toán học ý còn hỏi ở đây toàn đứa ngu hỏi làm gì

a) \(-x+80=-220-5x\)

\(\Rightarrow-x+5x=-220-80\)

\(\Rightarrow4x=-300\)

\(\Rightarrow x=-\dfrac{300}{4}\)

\(\Rightarrow x=-75\)

b) \(98+\left(x-12\right)+68=-80-x\)

\(\Rightarrow166+\left(x-12\right)=-80-x\)

\(\Rightarrow166+x-12=-80-x\)

\(\Rightarrow154+x=-80-x\)

\(\Rightarrow154+80=-x-x\)

\(\Rightarrow-2x=234\)

\(\Rightarrow x=-\dfrac{234}{2}\)

\(\Rightarrow x=-117\)

c) \(122+x-78=-55-4x-1\)

\(\Rightarrow44+x=-56-4x\)

\(\Rightarrow x+4x=-56-44\)

\(\Rightarrow5x=-100\)

\(\Rightarrow x=-\dfrac{100}{5}\)

\(\Rightarrow x=-20\)

d) \(663+9x=-x-37\)

\(\Rightarrow9x+x=-37-663\)

\(\Rightarrow10x=-700\)

\(\Rightarrow x=-\dfrac{700}{10}\)

\(\Rightarrow x=-70\)

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

\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(x-2\right)^3=0\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)

\(\Rightarrow x=2\)

b) \(3^{x-1}:81=3^3\)

\(\Rightarrow3^{x-1}:3^4=3^3\)

\(\Rightarrow3^{x-1-4}=3^3\)

\(\Rightarrow3^{x-5}=3^3\)

\(\Rightarrow x-5=3\)

\(\Rightarrow x=8\)

c) \(x^{13}=x\)

\(\Rightarrow x^{13}-x=0\)

\(\Rightarrow x\left(x^{12}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{12}=1\end{matrix}\right.\)

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

d) \(\left(x-9\right)^4=\left(x-9\right)^2\)

\(\Rightarrow\left(x-9\right)^2=x-9\)

\(\Rightarrow\left(x-9\right)^2-\left(x-9\right)=0\)

\(\Rightarrow\left(x-9\right)\left(x-9-1\right)=0\)

\(\Rightarrow\left(x-9\right)\left(x-10\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-9=0\\x-10=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)

a) \(-\left(-a+c-d\right)-\left(c-a+d\right)\)

\(=a-c+d-c+a-d\)

\(=2a-2c\)

b) \(-\left(a+b-c+d\right)+\left(a-b-c-d\right)\)

\(=-a-b+c-d+a-b-c-d\)

\(=-2b-2d\)

c) \(a\left(b-c-d\right)-a\left(b+c-d\right)\)

\(=ab-ac-ad-ab-ac+ad\)

\(=-2ac\)

d) \(\left(a-b\right)+\left(c-d\right)+\left(a+c\right)-\left(b+d\right)\)

\(=a-b+c-d+a+c-b-d\)

\(=2a-2b+2c-2d\)

h) \(\left(a-b\right)-\left(c-d\right)-\left(a+d\right)-\left(b+c\right)\)

\(=a-b-c+d-a-d-b-c\)

\(=\left(a-a\right)-\left(b+b\right)-\left(c+c\right)+\left(d+d\right)\)

\(=-2b-2c\)

i) \(-\left(-a+c-d\right)-\left(c-a+d\right)\)

\(=a-c+d-c+a-d\)

\(=\left(a+a\right)-\left(c+c\right)+\left(d+d\right)\)

\(=2a-2c\)

k) \(-\left(a+b-c+d\right)+\left(a-b-c-d\right)\)

\(=-a-b+c-d+a-b-c-d\)

\(=\left(-a+a\right)-\left(b+b\right)+\left(c-c\right)-\left(d+d\right)\)

\(=-2b-2d\)

n) \(-a\left(-b-c+d-e\right)-\left(-a\right)-\left(-b+c-d\right)\)

\(=ab+ac-ad+ae+a+b-c+d\)

Bài `4`

`1, 2y(x+2)-3x-6`

`=2y(x+2) -(3x+6)`

`=2y(x+2) -3(x+2)`

`=(x+2)(2y-3)`

`2, 3(x+4) -x^2-4x`

`=3(x+4)-(x^2+4x)`

`=3(x+4) -x(x+4)`

`=(x+3)(3-x)`

`3, 2(x+5) -x^2-5x`

`=2(x+5)-(x^2+5x)`

`=2(x+5)-x(x+5)`

`=(x+5)(2-x)`

`4, x^2 +6x-3(x+6)`

`= (x^2+6x) -3(x+6)`

`=x(x+6)-3(x+6)`

`=(x+6)(x-3)`

`5, x(x+y) -5x-5y`

`=x(x+y) -(5x+5y)`

`=x(x+y)-5(x+y)`

`=(x+y)(x-5)`

`6,x(x-y)+2x-2y`

`=x(x-y)+2(x-y)`

`=(x-y)(x+2`

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Em ko muốn bài giải vậy Thầy xóa đây. Em tự xem lại cách học của mình nhé!