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Ta có :
5x + 1 - ( 5x - x2 )
= 5x + 1 - 5x + x2
= x2 + 1
vì x2 \(\ge\)0 nên x2 + 1 > 0
Vậy đa thức trên không có nghiệm
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1, Thay x = 1/3 ; y = -1/5 ta được
\(=\dfrac{3.1}{9}-5\left(-\dfrac{1}{5}\right)+1=\dfrac{1}{3}+2=\dfrac{7}{3}\)
2, Thay x = -2 ; y = -1/2 ta được
\(=5.4\left(-\dfrac{1}{2}\right)+3\left(-2\right)\left(-\dfrac{1}{2}\right)-\dfrac{2\left(-2\right).1}{4}\)
\(=-10+3+1=-6\)
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I don't now
sorry
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\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\left(DK:x\ne-\frac{2}{5};x\ne-\frac{1}{5}\right)\)
\(\Rightarrow\left(2x+3\right)\left(10x+2\right)=\left(4x+5\right)\left(5x+2\right)\Rightarrow20x^2+34x+6=20x^2+33x+10\Rightarrow x=4\)(thoả mãn)
Vậy x = 4
\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)
\(\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)
\(\Leftrightarrow2x\left(10x+2\right)+3\left(10x+2\right)=5x\left(4x+5\right)\)
\(\Leftrightarrow20x^2+4x+20x+6=20x^2+25x+9x+10\)
\(\Leftrightarrow20x^2+4x+20x+6-\left(20x^2+25x+9x+10\right)=0\)\(\Rightarrow20x^2+24x+6-\left(20x^2+34x+10\right)=0\)
\(\Leftrightarrow-10x-4=0\)
\(\Leftrightarrow-10x=4\)
\(\Leftrightarrow x=-\frac{4}{10}\)
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=>(3x+2)(5x+1)=(5x+7)(3x-1)
(3x)(5x+1)+2(5x+1)=(5x)(3x-1)+7(3x-1)
15x2+3x+10x+2=15x2-5x+21x-7
(15x2-15x2)+(3x+10x+5x-21x)=-7-2
0-3x=-9
-3x=-9
x=(-9)/(-3)
x=3
\(\frac{3x+2}{5x+7}\)=\(\frac{3x-1}{5x+1}\)
\(\Rightarrow\)\(\frac{3x+2}{3x-1}\)= \(\frac{5x+7}{5x+1}\)\(\Rightarrow\)1+\(\frac{3}{3x-1}\)=1+\(\frac{6}{5x+1}\)
\(\Rightarrow\)\(\frac{3}{3x-1}\)= \(\frac{6}{5x+1}\)\(\Rightarrow\)3.(5x+1) = 6.(3x-1)
\(\Rightarrow\)15x+3 = 18x -6
\(\Rightarrow\)3x = 6 +3
\(\Rightarrow\)x =3
Ta có: 5x - |2 + x| = 3
=> |2 + x| = 5x - 3
\(\Rightarrow\orbr{\begin{cases}2+x=5x-3\\2+x=3-5x\end{cases}\Rightarrow\orbr{\begin{cases}4x=5\\6x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{6}\end{cases}}}\)
Vậy x = 5/4 , x = 1/6
5x-/2+x/=3
=> /2+x/=5x-3
ĐK: 5x-3 > 0=> 5x>3=> x>3/5
Ta có: /2+x/=5x-3
=> \(\orbr{\begin{cases}2+x=5x-3\\2+x=3-5x\end{cases}}\)
=>\(\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{6}\left(loại\right)\end{cases}}\)
Vậy phương trình có nghiện x = \(\frac{5}{4}\)