a) \(0,25x^3+x^2+x=0\)

\(\Leftrightarrow x\left(0,25x^2+x+1\right)=0\)

\(\Leftrightarrow x\left[\left(\frac{1}{2}x\right)^2+2\cdot\frac{1}{2}x\cdot1+1^2\right]=0\)

\(\Leftrightarrow x\left(\frac{1}{2}x+1\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)

Vậy....

b) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)

\(\Leftrightarrow\frac{2-x}{2007}-1+2=\frac{1-x}{2008}+1+\frac{-x}{2009}+1\)

\(\Leftrightarrow\frac{2-x+2007}{2007}=\frac{1-x+2008}{2008}+\frac{-x+2009}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}-\frac{2009-x}{2008}-\frac{2009-x}{2009}=0\)

\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)

\(\Rightarrow2009-x=0\)

\(\Leftrightarrow x=2009\)

Vậy....

\(2\left(a^2+b^2\right)=5ab\)

\(\Leftrightarrow2a^2+2b^2-5ab=0\)

\(\Leftrightarrow\left(2a^2-4ab\right)+\left(2b^2-ab\right)=0\)

\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)

\(a>b\Rightarrow2a>b\Leftrightarrow2a-b>0\)

\(\Rightarrow a=2b\)

\(Q=\frac{3\left(2b-b\right)}{2\left(2b+b\right)}=\frac{3b}{6b}=\frac{1}{2}\)

thank bạn nha

4 ( x + y ) = xy + 11

\(\Leftrightarrow\)4x + 4y - xy = 11

\(\Leftrightarrow\)x ( 4 - y ) - 16 + 4y = -5

\(\Leftrightarrow\)x ( 4 - y ) - 4 ( 4 - y ) = -5

\(\Leftrightarrow\)( x - 4 ) ( 4 - y ) = -5

lập bảng giá trị, tìm được x,y

Câu hỏi của NOO PHƯỚC THỊNH - Toán lớp 8 - Học toán với OnlineMath

>: nhấn vô coi câu tl rồi bt( câu b ý) 

0 nha bạn hihi

^{ssssssssssssssss_ssssss}s

đáp án

12+91-12-91=9

học tốt