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a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)
\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)
b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)
2.I3x - 1I + 1 = 5
<=>2.I3x - 1I = 5-1
<=>2.I3x - 1I =4
<=>I3x - 1I=2
=>Có 2 trường hợp
3x-1=2 =>3x=3 =>x=1
3x-1=-2 =>3x=1 =>x=1/3
Vậy x có 2 giá trị thỏa mãn là 1 và 1/3
Học tốt ^-^
\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(\frac{-1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\hept{\begin{cases}\frac{1}{7}x-\frac{2}{7}=0\\\frac{-1}{5}x+\frac{3}{5}=0\\\frac{1}{3}x+\frac{4}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=3\\x=-4\end{cases}}}\)
KL
b, \(\left|\frac{5}{3}x\right|=\left|\frac{-1}{6}\right|\)
\(\left|\frac{5}{3}x\right|=\frac{1}{6}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}x=\frac{1}{6}\\\frac{5}{3}x=\frac{-1}{6}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}\\x=\frac{-1}{10}\end{cases}}}\)
KL
c, \(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\left|\frac{-3}{4}\right|\)
\(\left|\frac{3}{4}x-\frac{3}{4}\right|-\frac{3}{4}=\frac{3}{4}\)
\(\Rightarrow\left|\frac{3}{4}x-\frac{3}{4}\right|=\frac{3}{2}\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\\frac{3}{4}x-\frac{3}{4}=\frac{-3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{-3}{4}\end{cases}}}\)
KL
làm câu b , bài 1 nhé
A =(ghi lại )
=> 2A=2+22+23+24+....+2100+2101
=> 2A - A = A = 2+22+23+24+....+2100+2101 -1 -2-22-23-....-2100
=>A = 2101-1 < 2101
Vậy A < B
Bài 1:
a) \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{19}{81.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
b) Ta có: \(A=2^0+2^1+...+2^{100}\)
\(\Rightarrow2A=2+2^2+...+2^{101}\)
\(\Rightarrow2A-A=2^{101}-2^0\)
\(\Rightarrow A=2^{201}-1< 2^{101}\)
\(\Rightarrow A< B\)
Vậy A < B
\(\Leftrightarrow6^x\cdot\dfrac{1}{6}+6^x\cdot36=6^7\cdot217\)
\(\Leftrightarrow6^x=1679616\)
hay x=8
\(A=3x^3y^4+4xy^3-8y^3+2021-3y^4x^3\)
\(\Rightarrow A=\left(3x^3y^4-3y^4x^3\right)+4xy^3-8y^3+2021\)
\(\Rightarrow A=4xy^3-8y^3+2021\)
Thay x = 2; y = -3 ta có:
\(A=4\cdot2\cdot\left(-3\right)^3-8\cdot\left(-3\right)^3+2021\)
\(\Rightarrow A=-216-\left(-216\right)+2021\)
\(\Rightarrow A=2021\)
~~ Chúc bạn học tốt ~~
a) | 9 + 7x | = 3 - 5x
\(\Rightarrow\orbr{\begin{cases}9+7x=3-5x\\9+7x=-\left(3-5x\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}7x+5x=3-9\\9+7x=-3+5x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}12x=-6\\7x-5x=-3-9\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\2x=-12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=-6\end{cases}}\)