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a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
a) \(5+3^{x+1}=86\)
\(=>3^{x+1}=86-5\)
\(=>3^{x+1}=81=3^4\)
\(=>x+1=4\) ( cùng cơ số )
\(=>x=4-1\)
\(=>x=3\)
b) \(15:\left(x+2\right)=\left(3^3+3\right):10\)
\(=>15:\left(x+2\right)=\left(27+3\right):10\)
\(=>15:\left(x+2\right)=30:10=3\)
\(=>x+2=15:3\)
\(=>x+2=5\)
\(=>x=5-2\)
\(=>x=3\)
c) \(\left(9x+2\right).4=80\)
\(=>9x+2=80:4\)
\(=>9x+2=20\)
\(=>9x=20-2\)
\(=>9x=18\)
\(=>x=18:9\)
\(=>x=2\)
d) \(\left(245-x\right)+7^2=14\)
\(=>\left(245-x\right)+14=14\)
\(=>245-x=14-14\)
\(=>245-x=0\)
\(=>x=245-0\)
\(=>x=245\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=1-\frac{1}{2020}\)
\(A=\frac{2019}{2020}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)
\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(2B=1-\frac{1}{2019}\)
\(2B=\frac{2018}{2019}\)
\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)
1,
a,-3/5
b,-1/2
c,19/39
d,1/4
e,-39/40
f,-59/56
2,
a,=
b,<
c,>
d,<
k cho mình nha
x(x+2)=0
suy ra x=0 hoặc x+2=0
5-2x=-7
2x=-7+5
2x=-(7-5)
2x=-2
x=-2:2
x=-1
Vậy x=-1
NHỚ TÍCH MK NHA
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Jim Rohn – Triết lý cuộc đời
\(a)13-\left(40-X\right)=35\)
\(\Leftrightarrow40-X=13-35\)
\(\Leftrightarrow40-X=-22\)
\(\Leftrightarrow X=40-\left(-22\right)\)
\(\Leftrightarrow X=62\)
Vậy \(X=62\)
\(b)14-3x\left(5-X\right)=8\)
\(\Leftrightarrow3x\left(5-X\right)=14-8\)
\(\Leftrightarrow3x\left(5-X\right)=6\)
\(\Leftrightarrow5-X=6:3\)
\(\Leftrightarrow5-X=3\)
\(\Leftrightarrow X=5-3\)
\(\Leftrightarrow X=2\)
Vậy \(X=2\)
\(c)\left(3X-2\right)x\left(5+X\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3X-2=0\\5+X=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3X=0+2\\X=0-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3X=2\\X=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}X=2:3\\X=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}X=\frac{2}{3}\\X=-5\end{cases}}\)
Vậy \(X\in\left\{-5;\frac{2}{3}\right\}\)
\(d)\left(3X-6\right)x3=3^4\)
\(\Leftrightarrow\left(3X-6\right)x3=81\)
\(\Leftrightarrow3X-6=81:3\)
\(\Leftrightarrow3X-6=27\)
\(\Leftrightarrow3X=27+6\)
\(\Leftrightarrow3X=33\)
\(\Leftrightarrow X=33:3\)
\(\Leftrightarrow X=11\)
Vậy\(X=11\)
Hả 2+b - b+ 10 là 3c há
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