\(\frac{5}{3}\) . \(8^{x+2}\)- \(\frac{3}{5}\). \(8^x\)= \(\frac{5}{3}\). \(8^{11}\)- \(\frac{3}{5}\). \(8^9\)
tìm x :
\(\frac{5}{8}.8^{x+2}-\frac{3}{5}.8^x=\frac{5}{3}.8^{11}-\frac{3}{5}.8^9\)
Đúng ghi Đ, sai ghi S vào chỗ chấm:
a) \(\frac{3}{7}+\frac{1}{3}x\frac{3}{7}=\frac{16}{21}x\frac{3}{7}=\frac{16}{49}\) ......
b)\(\frac{5}{9}+\frac{1}{5}x\frac{5}{6}=\frac{5}{9}x\frac{1}{6}=\frac{13}{18}\) ......
c)\(\frac{13}{9}-\frac{7}{9}:\frac{2}{3}=\frac{6}{9}:\frac{3}{2}=1\) ......
d)\(\frac{11}{8}-\frac{5}{8}:\frac{3}{4}=\frac{11}{8}-\frac{5}{8}x\frac{4}{3}=\frac{11}{8}-\frac{10}{24}=\frac{23}{24}\) ......
Mai Hồng Ngọc? Vũ Thị Thu Hằng? Ai đúng dzậy -_-*
\(5\frac{6}{11}x+8\frac{9}{11}x+2\frac{3}{11}=3\frac{4}{11}x-\frac{8}{11}\)
\(5\frac{6}{11}x+8\frac{9}{11}x+2\frac{3}{11}=3\frac{4}{11}x-\frac{8}{11}\)
\(\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\)
\(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
a) Đk: x \(\ne\)-2
Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)
<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)
<=> 2x2 - 4x + 8 - 2x2 - 16 = 5x + 10
<=> -4x - 8 = 5x + 10
<=> -4x - 5x = 10 + 8
<=> -9x = 18
<=> x = -2 (ktm)
=> pt vô nghiệm
b) Đk: x \(\ne\)2; x \(\ne\)-3
Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)
<=> x + 3 - 6x + 12 = -5
<=> -5x = -5 - 15
<=> -5x = -20
<=> x = 4
vậy S = {4}
c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11
Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)
<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)
<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)
<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)
<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)
Vậy S = {0}
x:\(\left(9\frac{1}{2}-\frac{3}{2}\right)\)=\(\frac{\frac{2}{9}+\frac{2}{5}-\frac{2}{11}}{\frac{8}{5}+\frac{8}{9}-\frac{8}{11}}\)
giải hộ mình với
\(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{\frac{2}{9}+\frac{2}{5}-\frac{2}{11}}{\frac{8}{5}+\frac{8}{9}-\frac{8}{11}}\)
\(x:8=\frac{2\left(\frac{1}{9}+\frac{1}{5}-\frac{1}{11}\right)}{8\left(\frac{1}{5}+\frac{1}{9}-\frac{1}{11}\right)}\)
\(x:8=\frac{1}{4}\)
\(x=2\)
Vậy..........
\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)
\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)
\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)
\(=\frac{5}{8}:\frac{15}{16}\)
\(=\frac{2}{3}\)
\(\frac{\frac{5}{12}+\frac{1}{8}-\frac{7}{11}}{\frac{49}{11}-\frac{7}{8}-\frac{35}{12}}\)
\(\frac{5-\frac{5}{3}-\frac{5}{9}-\frac{5}{27}}{-8+\frac{8}{3}+\frac{8}{9}+\frac{8}{27}}:\frac{15-\frac{15}{11}-\frac{15}{121}}{16-\frac{16}{11}-\frac{16}{121}}\)
tìm x biết
1, \(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)
2, \(\frac{x}{120}=\frac{3}{8}\cdot\frac{-4}{25}\)
3, \(\frac{11}{13}\cdot x=1\)
4, \(-\frac{9}{8}+\frac{-3}{8}\cdot x=\frac{-1}{8}\)
5, \(\frac{-2}{5}\cdot x+\frac{4}{3}=\frac{7}{3}\)
\(1,\)\(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)
\(x-\frac{3}{5}=\frac{3}{35}+\frac{7}{6}\)
\(x-\frac{3}{5}=\frac{263}{210}\)
\(x=\frac{263}{210}+\frac{3}{5}\)
\(x=\frac{389}{210}\)
VẬY: \(x=\frac{389}{210}\)
\(F=\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)
\(\frac{5\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\times\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}\div\frac{15\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\times\left(1-\frac{1}{11}+\frac{1}{121}\right)}=\frac{5}{8}\div\frac{15}{16}=\frac{2}{3}\)
F=5-5x(1/3+1/9-1/27) /8-8x(1/3+1/9-1/27)
: 15-15x(1/11+1/121) /16-16x(1/11+1/121)
=5-5x1/8-8x1
: 15-15x1/16-16x1
=0:0=0
chắc vậy!