Ôn tập: Phương trình bâc nhất một ẩn

a: y=2 thì \(A=\dfrac{x+2}{2-1}=x+2\)

\(B=\dfrac{4x\left(x+5\right)}{2+2}=x\left(x+5\right)\)

A+3=B

=>x+5=x(x+5)

=>(x+5)(1-x)=0

=>x=1 hoặc x=-5

b: Khi x=-3 thì \(A=\dfrac{-3+2}{y-1}=\dfrac{-1}{y-1}\)

\(B=\dfrac{4\cdot\left(-3\right)\cdot\left(-3+5\right)}{y+2}=\dfrac{-12\cdot2}{y+2}=\dfrac{-24}{y+2}\)

A-B=13

\(\Leftrightarrow-\dfrac{1}{y-1}+\dfrac{24}{y+2}=13\)

\(\Leftrightarrow13\left(y-1\right)\left(y+2\right)=-y-2+24y-24\)

\(\Leftrightarrow13y^2+13y-26=23y-26\)

=>y(13y-10)=0

=>y=0 hoặc y=10/13

a. đkxđ: x khác -2

\(1+\dfrac{1}{x+2}=\dfrac{12}{8+x^3}\)

\(\Leftrightarrow\dfrac{x^3+2^3}{\left(x+2\right)\left(x^2-2x+4\right)}+\dfrac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow x^3+8+x^2-2x+4=12\)

\(\Leftrightarrow x^3+x^2-2x=0\)

\(\Leftrightarrow x^3-x^2+2x^2-2x=0\)

\(\Leftrightarrow\left(x^3-x^2\right)+\left(2x^2-2x\right)=0\)

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

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

\(\Leftrightarrow x\left(x+2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\left(loai\right)\\x=1\end{matrix}\right.\)

Vậy...........

b.

\(\dfrac{4x-8}{2x^2+1}=0\)

\(\Leftrightarrow4x-8=0\)

\(\Leftrightarrow x=2\)

Vậy..........

Ngta kêu làm gì vậy bạn??

a)A= x²-10=-3x

⇔x²+3x-10=0

⇔x²+5x-2x-10=0

⇔(x²+5x)-(2x+10)=0

⇔x(x+5)-2(x+5)=0

⇔(x+5)(x-2)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

vậy pt A có tập n^o S={-5;2}

b)

B=\(\dfrac{A}{x^2+10}=\dfrac{x^2-10}{x^2+10}=\dfrac{x^2+10-20}{x^2+10}=1-\dfrac{20}{x^2+10}\)

Do \(x^2\ge0\forall x\)

=>\(x^2+10\ge10\)

=>\(\dfrac{20}{x^2+10}\le2\)

=>\(-\dfrac{20}{x^2+10}\ge-2\)

=>\(1-\dfrac{20}{x^2+10}\ge-1\)

=> B\(\ge-1\)

=> GTNN B=-1

Ta có:

\(\left(3a-b\right)^2=9a^2-6ab+b^2\)

\(\left(3a+b\right)^2=9a^2+6ab+b^2\)

\(\Rightarrow\left(3a-b\right)^2-\left(3a+b\right)^2=9a^2-6ab+b^2-\left(9a^2+6ab+b^2\right)\)

\(=9a^2-6ab+b^2-9a^2-6ab-b^2\)

\(=-12ab\)

=> ĐPCM

\(\text{Ta có : }VT=\left(3a-b\right)^2-\left(3a+b\right)^2\\ \\ =\left(3a-b+3a+b\right)\left(3a-b-3a-b\right)\\ \\ =6a\cdot\left(-2b\right)\\ =-12ab=VP\left(đpcm\right)\)

Vậy đảng thức được chứng minh.

\(\left|x+1\right|=\left|x\left(x+1\right)\right|\Rightarrow\left[{}\begin{matrix}x+1=x\left(x+1\right)\\x+1=-x\left(x+1\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\pm1\) Vậy \(S=\left\{\pm1\right\}\)

\(\left(4x+3\right)^2-4\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(4x+3\right)^2-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(4x+3\right)^2-\left(2x-2\right)^2=0\)

\(\Leftrightarrow\left(4x+3-2x+2\right)\left(4x+3+2x-2\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=0\\6x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}\\x=\dfrac{-1}{6}\end{matrix}\right.\)

Vậy .......................