(x nhân 11):3=(3372nhan 5):3
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a) 8/3 x 2/5 x 3/8 x 10 x 19/92
= ( 8/3 x 3/8 ) x ( 2/5 x 10 ) x 19/92
= 1 x 4 x 19/92
= 4 x 19/92 = 76/92 = 19/23
b) N = 5/7 x 5/11 + 5/7 x 2/11 - 5/7 x 14/11
= 5/7 x ( 5/11 + 2/11 - 14/11 )
= 5/7 x ( -7/11 )
= -5/11
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\(\frac{10}{3}.x+\frac{67}{4}=\frac{53}{4}\)
\(\frac{10}{3}.x=\frac{53}{4}-\frac{67}{4}\)
\(\frac{10}{3}.x=-\frac{7}{2}\)
\(x=-\frac{7}{2}:\frac{10}{3}\)
\(x=-\frac{21}{20}\)
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7(x-3)-5(3-x)=11x-5
<=>7(x-3)+5(x-3)=11x-5
<=>(7+5)(x-3)=11x-5
<=>12(x-3)=11x-5
<=>12x-36=11x-5
<=>12x-11x=-5+36
<=>x=31
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1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
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a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
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\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{11}{75}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{11}{75}:\frac{1}{2}=\frac{22}{75}\Leftrightarrow\frac{1}{x+2}=\frac{1}{25}\Leftrightarrow x=23\)
\(\left(x\times11\right):3=\left(3372\times5\right):3\)
\(\left(x\times11\right):3=16860:3\)
\(\left(x\times11\right):3=5620\)
\(x\times11=5620\times3\)
\(x\times11=16860\)
\(x=16860:11\)
\(x=\dfrac{16860}{11}\)