Giải phương trình sau:
\(\frac{7}{8}-\frac{1}{3}\cdot x=\frac{7}{10}+\frac{2}{3}\cdot x\)
Giải phương trình:
\(\frac{7}{10}-\frac{5}{8}\cdot x-\frac{7}{6}=\frac{5}{6}\cdot x\)
Ta có: \(\frac{7}{10}-\frac{5}{8}.x-\frac{7}{6}=\frac{5}{6}.x\)
\(\Rightarrow\frac{7}{10}-\frac{7}{6}=\frac{5}{6}x+\frac{5}{8}x\)
\(\Rightarrow\frac{21}{30}-\frac{35}{30}=\left(\frac{5}{6}+\frac{5}{8}\right)x\)
\(\Rightarrow-\frac{7}{15}=\left(\frac{15}{24}+\frac{20}{24}\right)x\)
\(\Rightarrow-\frac{7}{15}=\frac{35}{24}x\)
\(\Rightarrow x=-\frac{7}{15}:\frac{35}{24}\)
\(\Rightarrow x=-\frac{7}{15}.\frac{24}{35}\)
\(\Rightarrow x=-\frac{8}{25}\)
Vậy \(x=-\frac{8}{25}\)
Chuk bạn hok tốt!
Giải phương trình:
\(\frac{7}{10}-\frac{5}{8}\cdot x-\frac{7}{6}=\frac{5}{6}\cdot x\)
Giải phương trình
\(\left(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+\cdot\cdot\cdot+\frac{1}{10\cdot60}\right)\cdot x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+\cdot\cdot\cdot+\frac{1}{50\cdot60}\)
Giải phương trình
\(\left(\frac{1}{1\cdot51}+\frac{1}{2\cdot52}+\frac{1}{3\cdot53}+\cdot\cdot\cdot+\frac{1}{10\cdot60}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+\cdot\cdot\cdot+\frac{1}{50\cdot60}\)
Giải phương trình sau:
\(\frac{4}{5}-\frac{3}{2}\cdot x=\frac{7}{15}\)
\(\frac{4}{5}-\frac{3}{2}.x=\frac{7}{15}\)
\(\Leftrightarrow-\frac{3}{2}x=\frac{7}{15}-\frac{4}{5}=-\frac{1}{3}\)
\(\Leftrightarrow x=-\frac{1}{3}:\left(-\frac{3}{2}\right)=\frac{2}{9}\)
vậy x=\(\frac{2}{9}\)
\(\frac{4}{5}-\frac{3}{2}.x=\frac{7}{15}\)
\(\frac{3}{2}x=\frac{4}{5}-\frac{7}{15}\)
\(\frac{3}{2}x=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{3}{2}\)
\(x=\frac{2}{9}\)
Ta có: \(\frac{4}{5}-\frac{3}{2}.x=\frac{7}{15}\)
\(\Rightarrow\frac{3}{2}.x=\frac{4}{5}-\frac{7}{15}\)
\(\Rightarrow\frac{2}{3}.x=\frac{12}{15}-\frac{7}{15}\)
\(\Rightarrow\frac{2}{3}.x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}:\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{3}.\frac{3}{2}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
Chuk bn hok tốt!
Giải phương trình sau :
\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}\cdot\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\).
\(\text{GIẢI :}\)
ĐKXĐ : \(x\ne\pm1\)
\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}\cdot\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)
\(\Leftrightarrow\frac{2}{x+1}+\frac{x}{x-1}=\frac{x+1}{x^2-1}\)
\(\Leftrightarrow\frac{2}{x+1}+\frac{x}{x-1}-\frac{x+1}{x^2-1}=0\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{x+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow\text{ }2\left(x-1\right)+x\left(x+1\right)-(x+1)=0\)
\(\Leftrightarrow\text{ }2\left(x-1\right)+\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1\text{ (loại)}\\x=-3\text{ (Chọn)}\end{cases}}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-3\right\}\).
\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}.\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)\(đk:x\ne\pm1\)
\(< =>\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{\left[\frac{7}{6}.\frac{6}{7}+\left(1\right)\right]x+1}{x^2-1}\)
\(< =>\frac{2x-2+x^2+x}{x^2+x-x-1}=\frac{2x+1}{x^2-1}\)\(< =>\frac{x^2+3x-2}{x^2-1}=\frac{2x-1}{x^2-1}\)
\(< =>x^2+2x-2=2x-1\)\(< =>x^2+2x-2x-2+1=0\)
\(< =>x^2-1=0< =>x^2=1\)\(< =>x=\pm1\)\(\left(ktmđk\right)\)
Vậy phương trình trên vô nghiệm
tìm x
a) \(\frac{x-1}{2}+\frac{x-2}{5}=\frac{1}{4}+\frac{x-7}{10}\)
b) \(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{1}{2x-3}-\frac{3}{2}\)
c)\(7\cdot\left(x-1\right)+2x\cdot\left(1-x\right)=0\)
d) \(\frac{x+1}{2008}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2009}+\frac{x+11}{2008}+\frac{x+12}{2007}\)
e) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}+\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}+\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Tính nhanh giá trị biểu thức sau:
a) \(-\frac{9}{10}\cdot\frac{5}{14}+\frac{1}{10}\cdot\left(-\frac{9}{2}\right)+\frac{1}{7}\cdot\left(-\frac{9}{10}\right)\)
b)\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right)\cdot132\)
c)\(-\frac{2}{3}\cdot\left(\frac{8}{9}\cdot\frac{8}{13}-\frac{8}{27}\cdot\frac{3}{13}+\frac{4}{3}\cdot\frac{22}{39}\right)\)
a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)
= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)
= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)
b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)
\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)
\(=66+44+33+22+12=177\)
c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)
= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)
= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)
= còn lại làm nốt nha! bận ròy
\(\left(\frac{1}{7}\cdot x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}\cdot x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}\cdot x+\frac{4}{3}\right)=0\)
( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0
<=> +) 1/7 . x - 2/7 = 0 +) (- 1 / 5) . x +3/5 = 0 +) 1/ 3 . x + 4/ 3 = 0
x = 2 x = 3 x = 4
Vậy x = 2 : x = 3 ; x=4