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\(\text{a/-(59-3x)+39 = 3x -20}\)
\(-59+3x+39=3x-20\)
\(3x-59+39=3x-20\)
\(3x-20=3x-20\)
\(\Leftrightarrowđpcm\)
\(\text{b/ -(a+b+c) +(b-c) - (a-c-1)= 1+c- 2a }\)
\(-a-b-c+b-c-a+c+1=1+c-2a\)
\(1+\left(-b+b\right)+\left(-c-c+c\right)+\left(-a-a\right)=1+c-2a\)
\(1+0-c-2a=1+c-2a\)
\(1-c-2a=1+c-2a\)
\(\Rightarrow\)không thỏa mãn đề bàii
\(\text{c/ -(19-2x)=39= 2x+20}\)
\(-19+2x+39=2x+20\)
\(2x+\left(-19+39\right)=2x+20\)
\(2x+20=2x+20\)
\(\Leftrightarrowđpcm\)
\(\text{d/ b/ -(a+b+c) +(b-c) - (a-c+1)= c-2a-1 }\)
\(-a-b-c+b-c-a+c-1=c-2a-1\)
\(\left(-b+b\right)+\left(-c-c+c\right)+\left(-a-a\right)-1=c-2a-1\)
\(0-c-2a-1=c-2a-1\)
\(-c-2a-1=c-2a-1\)
\(\Leftrightarrow ko\)thỏa mãn đề bài
chúc bạn học tốt
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1 Tìm số nguyên x bik:
a.-( 2x+2)+25=-19
-2x - 2 + 25 = - 19
-2x = -19+2-25
-2x = -42
x = 21
b.1-(12+3x)=7
1 - 12 - 3x = 7
- 11 - 3x = 7
-3x = 7 + 11
- 3x = 18
x = -6
c.-(2x+2)+25=-19
giống câu a nhé
2.Rút gọn biểu thức
a.2x+(-61)-(21-61)
= 2x - 61 - 21 + 61
= 2x - 21
b. (-3-x+5)+3
= 2 - x + 3
= -1 -x
c.11-(13-x)+(13-11)
= 11 - 13 + x + 13 - 11
= x
d.25-(15-x+303)+303
= 25 - 15 + x - 303 + 303
= 10 + x
e.x+(-81)-(11-8)
= x - 81 - 11 + 8
= x - 84
f. (-1-x+2)+1
= -1 - x + 2 + 1
= - x + 2
g.15-(11-x)+(11-15)
= 15 - 11 + x + 11 - 15
= x
i) 15-(15-x+202)+202
= 15 - 15 + x - 202 + 202
= x
3.Chứng minh đẳng thức
a.-(59-3x)+39=3x-20
- 59 + 3x + 39 = 3x - 20
- 59 + 39 + 20 = 3x - 3x
0 = 0
b.-(a+b+c)+(b-c)-(a-c-1)=1+c-29
b - a - b - c + b - c - a + c + 1 = 1 + c - 29
b - a - b - c + b - c - a + c - c = 1 - 29 - 1
- 2a + b - 2c = - 29
chịu luôn -__-
c.-(19-2x)+39=2x+20
- 19 + 2x + 39 = 2x + 20
2x - 2x = 20 + 19 - 29
0 = 0
d.-(a+b+c)+b-c-( a-c+1)=c-2a-1
- a - b - c + b - c - a + c - 1 = c - 2a - 1
- a - b - c + b - c - a + c + 2a - c = -1 + 1
- 2c = 0
c = 0
Vậy c = 0
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gợi ý :Thực hiện phép tính ở từng vế
Tụ làm nốt
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Bài 1:
a) Ta có: \(\left(2x-1\right)^{20}=\left(2x-1\right)^{18}\)
\(\Leftrightarrow\left(2x-1\right)^{20}-\left(2x-1\right)^{18}=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^{18}\cdot\left(2x-2\right)\cdot2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
b) Ta có: \(\left(2x-3\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
c) Ta có: \(\left(x-5\right)^2=\left(1-3x\right)^2\)
\(\Leftrightarrow\left(x-5\right)^2-\left(3x-1\right)^2=0\)
\(\Leftrightarrow\left(x-5-3x+1\right)\left(x-5+3x-1\right)=0\)
\(\Leftrightarrow\left(-2x-4\right)\left(4x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Bài 2:
a) \(15^{20}-15^{19}=15^{19}\left(15-1\right)=15^{19}\cdot14⋮14\)
b) \(3^{20}+3^{21}+3^{22}=3^{20}\left(1+3+3^2\right)=3^{20}\cdot13⋮13\)
c) \(3+3^2+3^3+...+3^{2007}\)
\(=3\left(1+3+3^2\right)+...+3^{2005}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2005}\right)⋮13\)
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\(\text{VT=- 19+2x+39= 2x+20}\)
\(VT=-19 + 2x + 39 = 2x + 20\)