tìm x:
25+68:(5x+7)=29
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75 + 68 : (5x + 7) = 29
=> 68 : (5x + 7) = -46
=> 5x + 7 = -3128
=> 5x = -3135
=> x = -627
vậy_
(x - 7)(2x - 8) = 0
=> x - 7 = 0 hoặc 2x - 8 = 0
=> x = 7 hoặc x = 4
vậy_
(3x - 5)10 = (3x - 5)9
=> (3x - 5)10 - (3x - 5)9 = 0
=> (3x - 5)9 .[(3x - 5) - 1] = 0
=> (3x - 5)9 = 0 hoặc (3x - 5) - 1 = 0
=> 3x - 5 = 0 hoặc 3x - 5 = 1
=> 3x = 5 hoặc 3x = 6
=> x = 5/3 hoặc x = 2
vậy_
7x-x=521:519+3.22-70
=> 6x = 5^2 + 12 -1
=> 6x = 36
=> x = 36/6 = 6
Kết quả 6
Học tốt
a) Ta có: \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Leftrightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)
\(\Leftrightarrow\frac{x-17-33}{33}+\frac{x-21-29}{29}+\frac{x-2\cdot25}{25}=0\)
\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\)
nên x-50=0
hay x=50
Vậy: x=50
b) Ta có: \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)=2\)
\(\Leftrightarrow-15x^2+46x-35+15x^2-4x-4-2=0\)
\(\Leftrightarrow42x-41=0\)
\(\Leftrightarrow42x=41\)
hay \(x=\frac{41}{42}\)
a, \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=4-4\)
\(\Leftrightarrow\left(\frac{x-17-33}{33}\right)+\left(\frac{x-21-29}{29}\right)+\left(\frac{x-2.25}{25}\right)=0\)
\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\) (*)
Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\Rightarrow\) Phương trình (*) xảy ra khi: \(x-50=0\Leftrightarrow x=50\)
Vậy phương trình có nghiệm duy nhất là x = 50.
a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(\left(7\cdot x-21\right)\cdot x+25=29\)
=> \(7x^2-21x+25=29\)
=> \(7x^2-21x=4\)
=> \(7x\left(x-3\right)=4\)
=> \(x-3=\frac{4}{7x}\)
=> \(x=\frac{4}{7x}+3=\frac{4+21x}{7x}\)
Đề sai rồi nhá ...
Cái khúc " => " phần thứ 5 sửa lại là \(x\left(x-3\right)=\frac{4}{7}\),phần thứ 6 bỏ đi ...
Sai đề rồi nhé bạn , bạn sửa lại đi
1) 9-25=(7-x)-(25+7)
(7-x)-(25+7)=9-25
(7-x)-(25+7)=-16
(7-x)-32=-16
7-x=-16+32
7-x=16
x=7-16
x=-9
2) (27-514) -(486-73)+x=7
(27-514) -(486-73)+x=7
-487-413+x=7
-1011+x=7
x=7-(-1011)
x=7+1011
x=1018
3) 25+5+37-25+6-29-x=37
25+5+37-25+6-29-x=37
67-25+6-29-x=37
42+6-29-x=37
48-29-x=37
19-x=37
x=19-37
x=-18
4) 14+(-12)+x =10-/-15/+ /-3/
14+(-12)+x =10-15+3
14+(-12)+x =-5+3
14+(-12)+x =-2
2+x=-2
x=-2-2
x=-4
1) 5x + 3 = 2(x + 7)
5x + 3 = 2x + 2.7
5x + 3 = 2x + 14
5x - 2x = 14 - 3
3x = 11
x = 11/3
Vậy x = 11/3
2) 5x - 25 = 2x - 3
5x - 2x = -3 + 25
3x = 22
x = 22/3
Vậy x = 22/3
25 + 68 : ( 5x + 7 ) = 29
68 : ( 5x + 7 ) = 29 - 25
68 : ( 5x + 7 ) = 4
5x + 7 = 68 : 4
5x + 7 = 17
5x = 17 - 7
5x = 10
x = 10 : 5
x = 2
dap an la 2