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\(1,\left(\left(3\cdot\left(3x+2\right)\right)+4\cdot\left(x+2\right)\right)-90=0.\)
\(13x-76=0\)
\(13x=76\)
\(x=\frac{76}{13}\)
\(2,\left(\left(5\cdot\left(x+1\right)\right)+3\cdot\left(2x+1\right)\right)-65=0\)
\(11x-57=0\)
\(11x=57\)
\(x=\frac{57}{11}\)
\(3,\left(\left(3\cdot\left(3x+4\right)\right)+5\cdot\left(2x+1\right)\right)-100=0\)
\(19x-83=0\)
\(19x=83\)
\(x=\frac{83}{19}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{100}{609}\\ \)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{1}{3}-\frac{1}{2x+3}\)\(=\frac{2x}{3\left(2x+3\right)}\)
\(A=\frac{x}{3\left(2x+3\right)}=\frac{100}{609}=\frac{100}{3.203}=\frac{100}{3\left(2.100+3\right)}\)\(\Rightarrow x=100\)
a) \(4\left(\frac{1}{2}x-\frac{1}{3}\right)^2+5=\frac{61}{9}\)
=> \(4\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{61}{9}-5\)
=> \(4\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{16}{9}\)
=> \(\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{16}{9}:4\)
=> \(\left(\frac{1}{2}x-\frac{1}{3}\right)^2=\frac{16}{9\cdot4}=\frac{16}{36}=\frac{4}{9}\)
=> \(\frac{1}{2}x-\frac{1}{3}=\pm\frac{2}{3}\)
Trường hợp 1 : \(\frac{1}{2}x-\frac{1}{3}=\frac{2}{3}\)
=> \(\frac{1}{2}x=1\)
=> \(x=1:\frac{1}{2}=2\)
Trường hợp 2 : \(\frac{1}{2}x-\frac{1}{3}=-\frac{2}{3}\)
=> \(\frac{1}{2}x=-\frac{2}{3}+\frac{1}{3}=-\frac{1}{3}\)
=> \(x=\left(-\frac{1}{3}\right):\frac{1}{2}=\left(-\frac{1}{3}\right)\cdot2=-\frac{2}{3}\)
b) \(9\left(2x-\frac{1}{3}\right)^3-1=-\frac{2}{3}\)
=> \(9\left(2x-\frac{1}{3}\right)^3=-\frac{2}{3}+1=\frac{1}{3}\)
=> \(\left(2x-\frac{1}{3}\right)^3=\frac{1}{3}:9\)
=> \(\left(2x-\frac{1}{3}\right)^3=\frac{1}{3\cdot9}=\frac{1}{27}\)
=> \(2x-\frac{1}{3}=\frac{1}{3}\)
=> \(2x=\frac{2}{3}\)
=> \(x=\frac{2}{3}:2=\frac{1}{3}\)
Bài cuối tương tự
a)\(5^{2\text{x}-1}\)=174-49
\(^{5^{2\text{x}-1}}\)=125=\(^{5^3}\)
2x-1=3
2x=4
x=2
b)(3x+1)\(^3\)=10\(^{^{ }3}\)
3x+1=10
3x=9
x=3
Trả lời ;............................
a) x = 2
b) Sai đề.....................
Hk tốt..........................