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\(\frac{12+x}{43-x}=\frac{2}{3}\)\(\Rightarrow3\left(12+x\right)=2\left(43-x\right)\)
\(\Rightarrow36+3x=86-2x\)
\(\Rightarrow36+3x-86+2x=0\)
\(\Rightarrow5x=50\)
\(\Rightarrow x=10\)
\(\frac{12+x}{43-x}=\frac{2}{3}\)
\(\frac{\left(12+x\right)\times3}{\left(43-x\right)\times3}=\frac{2\times\left(43-x\right)}{3\times\left(43-x\right)}\)
\(\left(12+x\right)\times3=2\times\left(43-x\right)\)
\(36+x\times3=86-2\times x\)
\(x\times3+2\times x=86-36\)
\(x\times5=50\)
\(x=50\div5\)
\(x=10\)
Đặt \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
\(3A=3\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)
\(3A=3+1+...+\frac{1}{3^4}\)
\(3A-A=\left(3+1+...+\frac{1}{3^4}\right)-\left(1+\frac{1}{3}+...+\frac{1}{3^5}\right)\)
\(2A=3-\frac{1}{3^5}\)
\(A=\frac{3-\frac{1}{3^5}}{2}\)
Đặt \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\)
\(S=1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\)
\(S\times3=\left(1+\frac{1}{1\times3}+\frac{1}{3\times3}+\frac{1}{9\times3}+\frac{1}{27\times3}+\frac{1}{81\times3}\right)\times3\)
\(S\times3=3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\)
Xét: \(S\times3-S=\left(3+1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)-\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(S\times2=3-\frac{1}{243}\)
\(S\times2=\frac{728}{243}\)
\(S=\frac{728}{243}\div2\)
\(S=\frac{364}{243}\)
Vậy \(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{364}{243}\)
\(\left|x-\frac{1}{3}\right|+\left|x-y\right|=0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{3}=0\\x-y=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=\frac{1}{3}\\x=y\end{cases}\)\(\Leftrightarrow x=y=\frac{1}{3}\)
a.\(\frac{1}{6}.6^x+6^x.36=6^{15}\left(1+6^3\right)\)
\(6^x.\frac{217}{6}=6^{15}.217\)
\(6^x=6^{16}\)
\(x=16\)
Theo đề ta có:
\(\frac{a}{b}=\frac{a+6}{b+9}\)\(\Rightarrow a\left(b+9\right)=b\left(a+6\right)\)
\(\Rightarrow ab+9a=ab+6b\)
\(\Rightarrow ab+9a-ab-6b=0\)
\(\Rightarrow9x-6y=0\)
\(\Rightarrow9x=6y\Rightarrow\frac{x}{y}=\frac{6}{9}=\frac{2}{3}\)
Vậy phân số đó là \(\frac{2}{3}\)
Theo đề ta có:
\(\frac{a}{b}=\frac{a+6}{b+9}\Rightarrow a\left(b+9\right)=b\left(a+6\right)\)
\(\Rightarrow ab+9a=ab+6b\)
\(\Rightarrow ab+9a-ab-6b=0\)
\(\Rightarrow9a-6b=0\)
\(\Rightarrow9a=6b\Rightarrow\frac{a}{b}=\frac{6}{9}=\frac{2}{3}\)
Vậy phân số phải tìm là \(\frac{2}{3}\)
\(H=\left(9\frac{3}{8}+7\frac{3}{8}\right)+4,03=16\frac{3}{8}+4,03=16,375+4,03=20,405\)
\(I=10101.\left(\frac{5}{111111}+\frac{2,5}{111111}-\frac{4}{111111}\right)=10101.\frac{3,5}{111111}=\frac{7}{22}\)
a) Để A là 1 phân số thì
4 + n \(\ne\) 0
\(\Rightarrow\) n \(\ne\) - 4
b) A là 1 số nguyên
\(\Rightarrow\) n - 3
a) Để A là 1 phân số thì
4 + n ≠≠ 0
⇒⇒ n ≠≠ - 4
b) A là 1 số nguyên
⇒⇒ n - 3 chia hết cho n + 4
n +4 -7
a) Xét: \(1-\frac{3}{4}=\frac{1}{4}\); \(1-\frac{97}{98}=\frac{1}{98}\)
Vì \(\frac{1}{4}>\frac{1}{98}\) nên \(\frac{3}{4}< \frac{97}{98}\)
b) Xét: \(1-\frac{42}{43}=\frac{1}{43}\); \(1-\frac{112}{113}=\frac{1}{113}\)
Vì \(\frac{1}{43}>\frac{1}{113}\) nên \(\frac{42}{43}< \frac{112}{113}\)
Seo bn toàn ra những bài toán dễ thế ?