If \(\frac{x+24}{16}=\frac{45}{24}\)then \(x\)
Những câu hỏi liên quan
a. \(\frac{x+109}{3}+\frac{x+125}{5}+\frac{x+149}{7}+\frac{x+181}{9}=24\)24
b. \(\frac{x-96}{2}+\frac{x-88}{4}+\frac{x-76}{6}+\frac{x-60}{8}=14\)
c. \(\frac{155x-24}{51}+\frac{185x-48}{57}+\frac{205x-3}{61}\)= 16
Tìm x:
\(\left(x+\frac{1}{4}\right)^2+\frac{15}{16}=\frac{24}{16}\)
\(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x\frac{24}{25}x\frac{35}{36}x\frac{48}{49}x\frac{63}{64}\)
Tính giá trị biểu thức
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)
\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)
\(=\frac{1.9}{2.8}=\frac{9}{16}\)
Rút gọn phân thức
\(\frac{x^{24}+x^{20}+x^{16} +...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
Tính bằng 2 cách:
\(\frac{5}{24}\)x \(\frac{5}{12}\)x 24
(\(\frac{1}{4}\)+\(\frac{2}{3}\)) x \(\frac{4}{5}\)
\(\frac{3}{7}\)x \(\frac{16}{33}\) + \(\frac{16}{33}\) x \(\frac{4}{7}\)
\(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)
ĐKXĐ: x≠8
Ta có: \(\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}\)
\(\Leftrightarrow\frac{3}{2\left(x-8\right)}+\frac{3x-20}{x-8}+\frac{1}{8}-\frac{13x-102}{3\left(x-8\right)}=0\)
\(\Leftrightarrow\frac{9}{6\left(x-8\right)}+\frac{6\left(3x-20\right)}{6\left(x-8\right)}+\frac{6\left(x-8\right)}{48\left(x-8\right)}-\frac{2\left(13x-102\right)}{6\left(x-8\right)}=0\)
\(\Leftrightarrow9+6\left(3x-20\right)+6\left(x-8\right)-2\left(13x-102\right)=0\)
\(\Leftrightarrow9+18x-120+6x-48-26x+204=0\)
\(\Leftrightarrow45-2x=0\)
\(\Leftrightarrow2x=45\)
hay \(x=\frac{45}{2}\)(tm)
Vậy: \(x=\frac{45}{2}\)
Rút gọn biểu thức:\(\frac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\).
Ta nhận thấy mẫu của biểu thức trên là:
x26+x24+x22+...+x2+1=(x26+x22+...+x2)+(x24+x20+...+x4+1)
=x2(x24+x20+...+x16+...+1)+(x24+x20+...+x4+1)
=(x24+x20+...+1)(x2+1)
Như vậy\(\frac{x^{24}+x^{20}+x^{16}+...+1}{\left(x^{24}+x^{20}+...+1\right)\left(x^2+1\right)}\)=\(\frac{1}{x^2+1}\)
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học giỏi vclllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll.......
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Tìm B biết \(B=\frac{1+x^2+x^4+...+x^{22}+x^{24}+x^{26}}{1+x^4+x^8+...+x^{16}+x^{20}+x^{24}}\)
\(B=\frac{1+x^2+x^4+...+x^{26}}{1+x^4+x^8+...+x^{24}}\)
\(=\frac{\frac{\left(x^2-1\right)\left(1+x^2+x^4+...+x^{26}\right)}{x^2-1}}{\frac{\left(x^4-1\right)\left(1+x^4+x^8+...+x^{24}\right)}{x^4-1}}\)
\(=\frac{\frac{x^{28}-1}{x^2-1}}{\frac{x^{28}-1}{x^4-1}}=\frac{x^4-1}{x^2-1}=x^2+1\)
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Nếu x + \(\frac{3}{16}\)= \(-\frac{5}{24}\)thì x = ..........
x + 3/16 = -5/24
=> x = -5/24 - 3/16
=> x = -19/48
Vậy x = -19/48
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\(x+\frac{3}{16}=-\frac{5}{24}\)
\(x=-\frac{5}{24}-\frac{3}{16}\)
\(x=-\frac{19}{48}\)
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