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Tớ biết cách làm rồi. Đây là lời giải các bạn tham khảo nhé !
Đặt x ^ 2 = a ( a \(\ge\) 0), y = b (b \(\ge\) 0).
Ta có : \(\frac{a+b}{10}=\frac{a-2b}{7}\) và a ^ 2.b ^ 2 = 81.
\(\frac{a+b}{10}=\frac{a-2b}{7}=\frac{\left(a+b\right)-\left(a-2b\right)}{10-7}=\frac{3b}{3}=b\left(1\right)\)
\(\frac{a+b}{10}=\frac{a-2b}{7}=\frac{2a+2b}{20}=\frac{\left(2a+2b\right)+\left(a-2b\right)}{20+7}=\frac{3a}{27}=\frac{a}{9}\left(2\right)\)
Từ (1) và (2) suy ra \(\frac{a}{9}=b\Rightarrow a=9b\)
Do a ^ 2.b ^ 2 = 81 nên (9b) ^ 2.b ^ 2 = 81\(\Rightarrow81b^4=81\Rightarrow b^4=1\Rightarrow b=1\)( vì b\(\ge\) 0)
Suy ra a = 9.1 = 9
Ta có x ^ 2 = 9 và y ^ 2 = 1
Suy ra : x = \(\pm\) 3, y = \(\pm\) 1
Vậy x = 3 thì y = 1
hoặc x = -3 thì y =1
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=> \(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)
=> (x + 2)(\(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\)) = 0
=> x + 2 = 0 (vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\)= 0)
=> x = -2
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Tìm x hả bạn ?
a ) \(\left(3x+\frac{1}{4}\right)^3=-27\)
\(\left(3x+\frac{1}{4}\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+\frac{1}{4}=-3\)
\(\Rightarrow3x=-3-\frac{1}{4}=-\frac{13}{4}\)
\(\Rightarrow x=-\frac{13}{4}:3=-\frac{13}{12}\)
Vậy x = \(-\frac{13}{12}\)
cho em hoi câu này xin các anh chị:
10mux x+4y = 2013
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\(a.\)
\(1-\dfrac{1}{2}\left(\dfrac{3}{2}-2x\right)=4x-\dfrac{1}{4}\)
\(\Rightarrow1-\dfrac{3}{4}+x=4x-\dfrac{1}{4}\)
\(\Rightarrow1-\dfrac{3}{4}+\dfrac{1}{4}=4x-x\)
\(\Rightarrow3x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{6}\)
\(b.\)
\(x^{10}=1024\)
\(\Rightarrow x^{10}=2^{10}\)
\(\Rightarrow x=2\)
\(c.\)
\(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
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\(10.3^x=81\)
\(3^x=80:10\)
\(3^x=8\)
\(\Rightarrow3^x=\varnothing\)
\(20-\left(x-1\right)^3=\frac{1}{16}\)
\(\left(x-1\right)^3=20-\frac{1}{16}\)
\(\left(x-1\right)^3=\frac{320}{16}-\frac{1}{16}\)
\(\left(x-1\right)^3=\frac{319}{16}\)
\(\Rightarrow\left(x-1\right)^3=\varnothing\Leftrightarrow x=\varnothing\)
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Bài 1
A= \(\frac{81^{10}.3^{17}}{27^{10}.9^{13}}\)
= \(\frac{\left(3^4\right)^{10}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{13}}\)
= \(\frac{3^{40}.3^{17}}{3^{30}.3^{26}}\)
= \(\frac{3^{57}}{3^{56}}\)= 3
\(\frac{x}{7}\)= \(\frac{-12}{49}\)
=> 49x = (-12) x 7
=> 49x = -84
=> x= \(\frac{-12}{7}\)
\(\left(x+1\right)^2=\dfrac{81}{10}\\ =>x+1=\pm\sqrt{\dfrac{81}{10}}=\pm\dfrac{9}{\sqrt{10}}=\pm\dfrac{9\sqrt{10}}{10}\\ =>x=\dfrac{-10\pm9\sqrt{10}}{10}\)
\(\left(x+1\right)^2=\dfrac{81}{10}\)
<=>\(\left(x+1\right)=\pm\sqrt{\dfrac{81}{10}}\)
<=> \(\left[{}\begin{matrix}x=\dfrac{-10+9\sqrt{10}}{10}\\x=-\dfrac{10+9\sqrt{10}}{10}\end{matrix}\right.\)