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\((x - 3).(2y + 1) = 7\)
Ý của bạn là chỉ yc tìm mỗi vế của biến x ạ?
\(\left(x-3\right)\cdot\left(2y+1\right)\in\text{Ư}\left(7\right)=\left\{1;7;-1;-7\right\}\)
`\Rightarrow \text {TH1:} x - 3 = 1`
`\Rightarrow x = 1 + 3`
`\Rightarrow x = 4`
`\text {TH2:} x - 3 = 7`
`\Rightarrow x = 7 + 3`
`\Rightarrow x = 10`
`\text {TH3:} x - 3 = -1`
`\Rightarrow x = -1 + 3`
`\Rightarrow x = 2`
`\text {TH4:} x - 3 = -7`
`\Rightarrow x = -7 + 3`
`\Rightarrow x = -4`
Vậy, `x \in {-4; 4; 2; 10}`
ta có
trường hợp 1:(x-3)=7
x-3=7
x=7+3
x=10
x-3=7
x=7+3
x=10
trường hợp 2:(x-3)=1
x-3=1
x=1+3
x=4
Ta có: `B = 1 + 3 + 3^2 + ... + 3^1991`
`= (1 + 3 + 3^2) + (3^3 + 3^4 + 3^5) + ... + (3^1989 + 3^1990 + 3^1992)`
`= 13 + 3^3 (1 + 3 + 3^2) + ... + 3^1989 (1 + 3 + 3^2)`
`= 13 + 3^3 . 13 + ... + 3^1989 . 13`
`= 13 (1 + 3^3 + ... + 3^1989)`
Vì \(13\left(1+3^3+...+3^{1989}\right)⋮13\) nên \(B⋮13\)
`B = 1 + 3 + 3^2 + ... + 3^1991`
= (1 + 3^4) + (3 + 3^5) + ... + (3^1987 + 3^1991)`
`= 82 + 3 (1 + 3^4) + ... + 3^1987 (1 + 3^4)`
`= 82 + 3 . 82 + ... + 3^1987 . 82`
`= 82 (1 + 3 + ... + 3^1987)`
Vì \(82\left(1+3+...+3^{1987}\right)⋮41\) nên \(B⋮41\)
`C = 3 + 3^2 + 3^3 + ... + 3^1000`
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{997}+3^{998}+3^{999}+3^{1000}\right)\)
`= 120 + 3^4 (3 + 3^2 + 3^3 + 3^4) + ... + 3^996 (3 + 3^2 + 3^3 + 3^4)`
`= 120 + 3^4 . 120 + ... + 3^996 . 120`
`= 120 (1 + 3^4 + ... + 3^996)`
Vì \(120\left(1+3^4+...+3^{996}\right)⋮120\) nên \(C⋮120\)
Ta có: \(C=3+3^2+3^3+...+3^{1000}\)
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{997}+3^{998}+3^{999}+3^{1000}\right)\)
\(=120\left(1+3^5+...+3^{997}\right)⋮120\)(đpcm)
\(3^{x+4}=9^{2x-1}\)
\(\Rightarrow3^{x+4}=3^{4x-2}\)
\(\Rightarrow x+4=4x-2\)
\(\Rightarrow3x=6\Rightarrow x=2\)
Bài 1:
\(a)\left(x+\dfrac{2}{3}\right)^3=\dfrac{125}{64}.\\ \Leftrightarrow\left(x+\dfrac{2}{3}\right)^3=\left(\dfrac{5}{4}\right)^3.\\ \Rightarrow x+\dfrac{2}{3}=\dfrac{5}{4}.\\ \Leftrightarrow x=\dfrac{7}{12}.\)
\(b)\left(x-\dfrac{1}{2}\right)^3=\dfrac{8}{343}.\\\Leftrightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{7}\right) ^3.\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{2}{7}.\\ \Leftrightarrow x=\dfrac{11}{14}.\)
Bài 2:
\(a)\left(x-\dfrac{1}{3}\right)^2=\dfrac{25}{9}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{5}{3}\right)^2.\\\left(x-\dfrac{1}{3}\right)^2=\left(\dfrac{-5}{3}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{5}{3}.\\x-\dfrac{1}{3}=\dfrac{-5}{3}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=\dfrac{-4}{3}.\end{matrix}\right.\)
\(b)\left(x-\dfrac{3}{4}\right)^2=\dfrac{49}{16}.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{7}{4}\right)^2.\\\left(x-\dfrac{3}{4}\right)^2=\left(\dfrac{-7}{4}\right)^2.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{7}{4}.\\x-\dfrac{3}{4}=\dfrac{-7}{4}.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}.\\x=-1.\end{matrix}\right.\)
II: Tự luận
Câu 4:
a) Để A là phân số thì \(2n-4\ne0\)
\(\Leftrightarrow2n\ne4\)
\(\Leftrightarrow n\ne2\)
b) Để A là số nguyên thì \(2n+2⋮2n-4\)
\(\Leftrightarrow2n-4+6⋮2n-4\)
mà \(2n-4⋮2n-4\)
nên \(6⋮2n-4\)
\(\Leftrightarrow2n-4\inƯ\left(6\right)\)
\(\Leftrightarrow2n-4\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
\(\Leftrightarrow2n\in\left\{5;3;6;2;7;1;10;-2\right\}\)
hay \(n\in\left\{\dfrac{5}{2};\dfrac{3}{2};3;1;\dfrac{7}{2};\dfrac{1}{2};5;-1\right\}\)
I trắc nghiệm
1.d ; 2.b; 3.a; 4.d;5.c; 6.a;7.d; 8.d;9.c;10.a;11.c;12.b
II tự luận
câu 1
a, 3/5+-2/5=1/5
b, (4/5+1/2)(3/13-8/13)=13/10*(-5/13)=-1/2
c, -5/7*2/11+(-5/7)*9/11+1=-5/7(2/11+9/11)+1=-5/7*1+1=-5/7+7/7=2/7
câu 2
a, x-(-5/120=-7/12
x=-7/12+(-5/12)
x= -1
vậy ...
b, x/20=7/10+(-13/20)
x/20=1/20
x=1
vậy ...
câu 3 tự vẽ hình
ta có xOy+tOy=tOx
thay số: 35+toy=70
tOy=35
-Oy là tia pg của xOt
b) \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)
\(\Rightarrow5^{2x-3}=5^2\cdot3+5^2\cdot2\)
\(\Rightarrow5^{2x-3}=5^2\cdot5\)
\(\Rightarrow5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=\dfrac{6}{2}\)
\(\Rightarrow x=3\)
f) \(30-\left[4\left(x-2\right)+15\right]=3\)
\(\Rightarrow4\left(x-2\right)+15=30-3\)
\(\Rightarrow4\left(x-2\right)+15=27\)
\(\Rightarrow4\left(x-2\right)=12\)
\(\Rightarrow x-2=3\)
\(\Rightarrow x=2+3\)
\(\Rightarrow x=5\)
h) \(740:\left(x+10\right)=10^2-2\cdot13\)
\(\Rightarrow740:\left(x+10\right)=100-26\)
\(\Rightarrow740:\left(x+10\right)=74\)
\(\Rightarrow x+10=740:74\)
\(\Rightarrow x+10=10\)
\(\Rightarrow x=0\)
k) \(\left[\left(6x-39\right):7\right].4=12\)
\(\Rightarrow\left(6x-39\right):7=3\)
\(\Rightarrow6x-39=3\cdot7\)
\(\Rightarrow6x-39=21\)
\(\Rightarrow6x=60\)
\(\Rightarrow x=\dfrac{60}{6}\)
\(\Rightarrow x=10\)
b: \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)
=>\(5^{2x-3}=5^2\cdot3+5^2\cdot2=5^2\cdot5=5^3\)
=>2x-3=3
=>2x=6
=>x=3
d:
\(5^{2x-3}-2\cdot5^2=5^2\cdot3\)
=>\(5^{2x-3}=5^2\cdot3+5^2\cdot2=5^2\cdot5=5^3\)
=>2x-3=3
=>2x=6
=>x=3
f: \(30-\left[4\left(x-2\right)+15\right]=3\)
=>4(x-2)+15=30-3=27
=>4(x-2)=12
=>x-2=3
=>x=3+2=5
h: \(\dfrac{740}{x+10}=10^2-2\cdot13\)
=>\(\dfrac{740}{x+10}=100-26=74\)
=>x+10=10
=>x=0
k: \(\left[\dfrac{\left(6x-39\right)}{7}\right]\cdot4=12\)
=>\(\dfrac{\left(6x-39\right)}{7}=3\)
=>6x-39=3*7=21
=>6x=60
=>x=60/6=10