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a ) \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{\left(-11\right)}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{19}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\frac{19}{70}=\frac{3}{35}\)
=> \(\frac{2}{5}+x+\frac{3}{2}=\frac{3}{7}-\frac{3}{35}=\frac{12}{35}\)
=> \(\frac{2}{5}+x=\frac{12}{35}-\frac{3}{2}=-\frac{81}{70}\)
=> \(x=-\frac{81}{70}-\frac{2}{5}=-\frac{109}{70}\)
b) \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
=> \(\frac{3}{4}x-6=\frac{5}{2}\)
=> \(\frac{3}{4}x=\frac{17}{2}\)
=> \(x=\frac{17}{2}:\frac{3}{4}=\frac{34}{3}\)
Câu c,d tự làm nhé
a. \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{-11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\left(\frac{19}{10}+x\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\frac{19}{10}-x=\frac{5}{14}-\left|\frac{19}{70}\right|=\frac{5}{14}-\frac{19}{70}\)
\(\Rightarrow-\frac{103}{70}-x=\frac{3}{35}\)
\(\Rightarrow x=-\frac{103}{70}-\frac{3}{35}\)
\(\Rightarrow x=-\frac{109}{70}\)
b. \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
\(\Rightarrow\frac{3}{4}\left(x-8\right)=\frac{5}{7}.\frac{7}{2}=\frac{5}{2}\)
\(\Rightarrow x-8=\frac{10}{3}\)
\(\Rightarrow x=\frac{34}{3}\)
c. \(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
\(\Rightarrow\frac{1}{2}=\frac{2}{3}-7x-4x=\frac{2}{3}-11x\)
\(\Rightarrow11x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{66}\)
d. \(4\left(\frac{1}{2}-x\right)-5\left(x-\frac{3}{10}\right)=\frac{7}{4}\)
\(\Rightarrow2-4x-5x+\frac{3}{2}=\frac{7}{4}\)
\(\Rightarrow2-9x=\frac{1}{4}\)
\(\Rightarrow9x=\frac{7}{4}\)
\(\Rightarrow x=\frac{7}{36}\)
ta có h(x)=\(\left(-8x^3+8x^3\right)+\left(3x^7-x^7-2x^7\right)+x^4-36+49\)
(=)h(x)=\(x^4+13\)
=>\(x^4+13=1\left(=\right)x^4=-12\)=> ko tồn tại x thỏa mãn
ta có \(x^4\ge0\)=>\(x^4+13\ge13>0\)
Vậy h(x)luôn nhận giá trị dương
\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)
\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)
\(=2\)
a) x(x-8)-x(x+1)=2
x2 -8x -x2-x=2
-9x=2
\(x=-\frac{2}{9}\)
b) (x+3)5 - 7(x+9)=0
5x + 15 -7x -63=0
-2x - 48 =0
-2x=48
x=-24
c)4(x-7)+7(x-2)=11
4x -28 + 7x -14=11
11x -42=11
11x=11+42
11x=53
x=\(\frac{53}{11}\)
a) \(\left|-\frac{2}{11}+\frac{3}{22}x\right|-\frac{1}{2}=\frac{5}{7}\)
=> \(\left|-\frac{2}{11}+\frac{3}{22}x\right|=\frac{17}{14}\)
=> \(\orbr{\begin{cases}-\frac{2}{11}+\frac{3}{22}x=\frac{17}{14}\\-\frac{2}{11}+\frac{3}{22}x=-\frac{17}{14}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{215}{21}\\x=-\frac{53}{7}\end{cases}}\)
b) \(-\frac{7}{8}x-5\frac{3}{4}=3\)
=> \(-\frac{7}{8}x-\frac{23}{4}=3\)
=> \(-\frac{7}{8}x=3+\frac{23}{4}=\frac{35}{4}\)
=> \(x=\frac{35}{4}:\left(-\frac{7}{8}\right)=\frac{35}{4}\cdot\left(-\frac{8}{7}\right)=-10\)
c) \(2x+\left(-\frac{2}{7}\right)-7=-11\)
=> \(2x-\frac{2}{7}-7=-11\)
=> \(2x=-11+7+\frac{2}{7}=-\frac{26}{7}\)
=> \(x=\left(-\frac{26}{7}\right):2=-\frac{13}{7}\)
d) \(\frac{3}{7}+x:\frac{14}{15}=\frac{1}{2}\)
=> \(x:\frac{14}{15}=\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\)
=> \(x=\frac{1}{14}\cdot\frac{14}{15}=\frac{1}{15}\)
Answer:
\(\left(x-7\right)x+1-\left(x-7\right)x+11=0\)
\(\Rightarrow x^2-7x+1-x^2+7x+11=0\)
\(\Rightarrow0x+12=0\)
\(\Rightarrow0x=-12\) (Vô lý)
\(\Rightarrow x\in\varnothing\)
Vậy ta chọn đáp án D.