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a) \(\dfrac{x}{48}=-\dfrac{4}{7}\Rightarrow x=-\dfrac{192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\Rightarrow x+\dfrac{4}{5}=1\)
\(\Rightarrow x=\dfrac{1}{5}\)
c) \(2\left|x-1\right|^2=72\Rightarrow\left|x-1\right|^2=36\)
\(\Rightarrow\left|x-1\right|=6\)
TH1: x - 1 = -6 => x = -5
TH2: x - 1 = 6 => x = 7
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\Rightarrow x=2\)
f) | x - 2 | = 1 + 4 = 5
TH1: x - 2 = -5 => x = -3
TH2: x - 2 = 5 => x = 7
a) \(\dfrac{x}{48}=\dfrac{-4}{7}\)
⇒ x.7=48.(-4)
7x = -192
x=\(\dfrac{-192}{7}\) Vậy x=\(\dfrac{-192}{7}\)
b) \(\left(x+\dfrac{4}{5}\right)-\dfrac{2}{5}=\dfrac{3}{5}\)
\(\left(x+\dfrac{4}{5}\right)=\dfrac{3}{5}+\dfrac{2}{5}\)
\(x+\dfrac{4}{5}=1\)
\(x=1-\dfrac{4}{5}\)
\(x=\dfrac{1}{5}\)
c) chưa từng gặp dạng với giá trị tuyệt đối sory
d) \(\dfrac{1}{6}x-\dfrac{2}{3}=2\)
\(\dfrac{1}{6}x=2+\dfrac{2}{3}\)
\(\dfrac{1}{6}x=\dfrac{8}{3}\)
\(x=\dfrac{8}{3}:\dfrac{1}{6}\)
\(x=16\)
e) \(\dfrac{x}{2,5}=\dfrac{4}{5}\)
=> x.5 = 4.2,5
5x=10
x=10:5
x=2
f) |x-2|-4=1
|x-2|=1+4
|x-2|=5
=>\(\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=5+2\\x=-5+2\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
đôi khi cũng có sai sót , hãy xem lại thật kĩ
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1: \(5\cdot3^x=5\cdot3^4\)
nên \(3^x=3^4\)
hay x=4
2: \(7\cdot4^x=7\cdot4^3\)
nên \(4^x=4^3\)
hay x=3
3: \(8\cdot7^x=8\cdot7^6\)
nên \(7^x=7^6\)
hay x=6
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a,(=)\(3^{x+1}.\left(3+4\right)=7.3^6\)
(=)\(3^{x+1}=3^6\)
=>x+1=6(=)x=5
b
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Bài 1:
\(A=\left(x^3.x^3.x^2\right).\left(y.y^4\right).\left(\frac{2}{5}.\frac{-5}{4}\right)\)
\(A=x^8.y^5.\left(-\frac{1}{2}\right)\)
\(B=\left(x^5.x.x^2\right).\left(y^4.y^2.y\right).\left(\frac{-3}{4}.\frac{-8}{9}\right)\)
\(B=x^8.y^7.\frac{2}{3}\)
Bài 2:
\(A=\left(15.x^2.y^3-12.x^2.y^3\right)+\left(11x^3.y^2-8.x^3.y^2\right)+\left(7x^2-12x^2\right)\)
\(A=3.x^2.y^3+2.x^3.y^2-5x^2\)
B tương tự nhé, đáp án là (theo mình)
\(B=\frac{5}{2}.x^5.y+\frac{7}{3}.x.y^4-\frac{1}{4}.x^2.y^3\)
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Bài 1 :
a) \(-3+\left(-4\right)-\left(-3\right)+\left(2+7-10\right)=-3-4+3+2+7-10=-5\)
b) \(3-\left(-3+2-7\right)+\left(-4\right)=3+3-2+7-4=7\)
c) \(7+\left(-2-3+7\right)-\left(-2\right)=7-2-3+7+2=17\)
d) \(-\left(-3\right)-\left(-2+3-8\right)+\left(-6\right)=3+2-3+8-6=4\)
Bài 2 :
a) \(x^2-2x-\left(3x-2x\right)=x^2-2x-3x+2x=x^2-3x\)
b) \(-\left(x^2+3x^2\right)-\left(-5x^2+3x\right)=-x^2-3x^2+5x^2-3x=x^2-3x\)
c) \(\left(x-y\right)-\left(x+3y+1\right)=x-y-x-3y-1=-4y-1\)
Bài 1:
a, -3+ (-4) - (-3) + (2 + 7 - 10)
= -3 - 4 + 3 + 2 + 7 - 10
= 5 - 10
= -5.
b, 3 - (-3 + 2 - 7) + (-4)
= 3 + 3 - 2 + 7 - 4
= 11 - 4
= 7
c, 7 + (-2 - 3 + 7) - (-2)
= 7 - 2 - 3 + 7 + 2
= 9 + 2
= 11.
d, - (-3) - (-2 + 3 - 8) + (-6)
= 3 + 2 - 3 + 8 - 6
= 10 - 6
= 4.
Mình chỉ làm bài 1 thôi nhé.
Chúc bạn học tốt!
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BT1: \(\left(3^2\right)^2-\left(-2^3\right)^2-\left(-5^2\right)^2=81-64-625=-608\)
BT2: a, \(\dfrac{1}{9}.27^x=3^x\)
\(3^{3x-2}=3^x\)
\(\Rightarrow3x-2=x\Rightarrow x=\dfrac{1}{2}\)
b, \(3^{-2}.3^4.3^x=3^7\)
\(3^{2+x}=3^7\Rightarrow2+x=7\)
\(\Rightarrow x=5\)
c, \(2^{-1}.2^x+4.2^x=9.2^5\)
\(2^x\left(2^{-1}+4\right)=288\)
\(\Rightarrow2^x=288:4,5=64=2^6\)
\(\Rightarrow x=6\)
d, \(\left(2x-3\right)^2=16=4^2\)
\(\Rightarrow2x-3=4\Rightarrow x=\dfrac{7}{2}\)
e, \(\left(3x-2\right)^5=-243=-3^5\)
\(\Rightarrow3x-2=-3\Rightarrow x=\dfrac{-1}{3}.\)
BT1: \(a,3^2.\dfrac{1}{243}.81^2.\dfrac{1}{33}=3^2.3^{-5}.3^8.3^{-1}\dfrac{1}{11}\)
\(=3^4.\dfrac{1}{11}=\dfrac{81}{11}\)
b, \(\left(4.5^3\right):\left(2^3.\dfrac{1}{10}\right)=100.5.\dfrac{1}{8}.10=625\)
thui tyuwj lamf ddi
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a: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
\(\Leftrightarrow2^x=2^{10}\cdot5:\dfrac{5}{2}=2^{10}\cdot5\cdot\dfrac{2}{5}=2^{11}\)
=>x=11
b: \(\Leftrightarrow3^x\cdot\dfrac{1}{3}+3^x\cdot9=3^{13}\cdot28\)
\(\Leftrightarrow3^x=3^{13}\cdot28:\dfrac{28}{3}=3^{14}\)
hay x=14
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Bài 1c)
\(\frac{1}{3}+x=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\)
\(\frac{1}{3}+x=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)
\(\frac{1}{3}+x=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(\frac{1}{3}+x=\frac{1}{3}-\frac{1}{11}=\frac{11}{33}-\frac{3}{33}=\frac{8}{33}\)
\(x=\frac{8}{33}-\frac{1}{3}=\frac{8}{33}-\frac{11}{33}=\frac{-3}{33}=\frac{-1}{11}\)
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Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
\(3^{x+2}-3^x=72\)
\(\Rightarrow3^x\left(9-1\right)=72\)
\(\Rightarrow3^x=72:8=9\)
\(\Rightarrow x=2\)
3x + 2 - 3x = 72
=> 3x .9 - 3x = 72
=> 3x.(9 - 1) = 72
=> 3x . 8 = 72
=> 3x = 72 : 8
=> 3x = 9
=> 3x = 32
=> x = 2