Tìm số nguyên x:
a) \(11-\left(15-49\right)=-x+(9-25)\)
b) \(x-+\left(-479-479\right)=-\left(26-74\right)\)
c) \(-\left(1930+2020\right)-x=\left(-96\right)+\left(-134\right)\)
Những câu hỏi liên quan
Tìm x:a) 3xleft(3x-8right)-9x^2+80b)6x-15-xleft(5-2xright)0c) x^3-16x0d) 2x^2+3x-50e) 3x^2-xleft(3x-6right)36f) left(x+2right)^2-left(x-5right)left(x+1right)17g) left(x-4right)^2-xleft(x+6right)9h) 4xleft(x-1000right)-x+10000i) x^2-360j) x^2y-2+x+x^2-2y+xy0k) xleft(x+1right)-left(x-1right).left(2x-3right)0l) 3x^3-27x0
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Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
tìm x biết
\(\frac{\left(2019-x^2\right)+\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)+\left(x-2020^2\right)}\) = \(\frac{19}{49}\)
Tìm x biết \(\frac{\left(2019-x\right)^2+\left(2019-x\right)\left(x-2020\right)}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)}\)\(\frac{+\left(x-2020\right)^2}{+\left(x-2020\right)^2}\)\(=\frac{19}{49}\)
tìm x biết
a.\(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)6\left(x+1\right)^2=49\)49
b.\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=25\)
c.\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
1) tìm x biết
a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)
b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)
c)\(\left(x-\frac{1}{2}\right)^{50}-\left(\frac{1}{3}\right)^{20}\times\left(-\frac{1}{9}\right)^{15}\)
2) chứng tỏ
\(\left(74^{m+1}+74^m\right)⋮25\)
a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2^2}{3^2}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{100}\)
\(\Rightarrow x=100\)
Vậy x = 100
b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\frac{2}{3}-x=\frac{1}{6}\)
\(\Rightarrow x=\frac{2}{3}-\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
2)
Ta có:
\(74^{m+1}+74^m=74^m.74^1+74^m=74^m.\left(74+1\right)=74^m.75⋮25\)
( vì \(75⋮25\) )
\(\Rightarrowđpcm\)
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chứng minh rằng các biểu thức sau không phụ thuộc vào x:
a. \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
b. \(B=\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)
c. \(C=x\left(x^3+x^2-3x-2\right)-\left(x^2-2\right)\left(x^2+x-1\right)\)
Tìm x, biết:a) left(5x+1right)^2dfrac{36}{49}b) left[left(-0,5right)^3right]^xdfrac{1}{64}c) 2020^{left(x-2right).left(2x+3right)}1d) left(x+1right)^{x+10}left(x+1right)^{x+4} với xin Ze) dfrac{3}{4}sqrt{x}-dfrac{1}{2}dfrac{1}{3}
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Tìm x, biết:
a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
b) \(\left[\left(-0,5\right)^3\right]^x=\dfrac{1}{64}\)
c) \(2020^{\left(x-2\right).\left(2x+3\right)}=1\)
d) \(\left(x+1\right)^{x+10}=\left(x+1\right)^{x+4}\) với \(x\in Z\)
e) \(\dfrac{3}{4}\sqrt{x}-\dfrac{1}{2}=\dfrac{1}{3}\)
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
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Tìm x, biết:
\(\frac{\left(2019-x\right)^2+\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}{\left(2019-x\right)^2-\left(2019-x\right)\left(x-2020\right)+\left(x-2020\right)^2}=\frac{19}{49}\)
Các bạn mong giúp mình sớm nhé
ủa bạn j ơi chữ x chành bành ra trên đề kìa mà bạn bảo tìm làm j nữa
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Tham khảo tại: Câu hỏi của Lương Đức Hưng - Toán lớp 8 | Học trực tuyến
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27 tháng 8 2023 lúc 20:19
Tìm x:
a) \(\dfrac{1}{3}.x+\dfrac{2}{5}\left(x-1\right)=0\)
b)\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=x\)
c)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
d)\(9.\left(3x+1\right)^2=16\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
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