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\(2x^4-x^3+2x^2+1=2x^4-2x^3+2x^2+x^3-x^2+x+x^2-x+1\\ \)
\(=2x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(2x^2+x+1\right)\)
Vậy a = 2; b = 1; c = 1.
a) Ta có : ( x+3 ).( x- 5 ) = 0
suy ra: x+3 = 0 hoặc x - 5 = 0
suy ra : x = -3 hoặc x = 5
KL : Vậy x = -3 hoặc x = 5
a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)
\(x+\frac{2}{5}=\frac{-7}{20}\)
\(x=\frac{-13}{20}\)
Vậy \(x=\frac{-13}{20}\)
b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)
\(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x=\frac{15}{8}\)
\(x=\frac{-15}{4}\)
Vậy \(x=\frac{-15}{4}\)
1.
2|x-6|+7x-2=|x-6|+7x
2|x-6| - |x-6|=7x-(7x-2)
|x-6| = 2
=>x-6 = +2
*x-6=2 *x-6 = -2
x =2+6 x = (-2)+6
x =8 x = 4
2.
|x-5|-7(x+4)=5-7x
|x-5|-7x-28 =5-7x
|x-5|-28 =5-7x+7x
|x-5|-28 = 5
|x-5| = 5+28
|x-5| = 33
=>x-5 = +33
*x-5=33 *x-5=-33
x =38 x = -28
3.
3|x+4|-2(x-1)=7-2x
3|x+4|-2x+2 =7-2x
3|x+4|-2 =7-2x+2x
3|x+4|-2 =7
3|x+4| =7+2
3|x+4| = 9
|x+4| =9:3
|x+4| = 3
=>x+4 = +3
*x+4=3 *x+4=-3
x =-1 x = -7
a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)
\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{-2}{3}\)
b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)
\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)
\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{-11}{20}\)
c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)
\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)
\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{-4}{5}\)
d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)
\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{2}{9}\)
e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)
\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)
\(\Rightarrow x=\dfrac{-7}{2}\)
\(a,50\%x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)
\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{3}{2}x=1\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)
\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)
\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{25}{4}\)
Bài lớp \(6\) chưa sử dụng dấu \(\Leftrightarrow\) chị nhé ! Vẫn phải sử dụng dấu \(\Rightarrow\) Khi nào bài lớp \(8\) trở lên thì cj hãy dùng \(\Leftrightarrow\) ạ