Ôn tập: Phương trình bâc nhất một ẩn

Gọi hình thoi đề bài cho là hình thoi ABCD, M,N,P,Q lần lượt là trung điểm của AB,AD,CD,BC

Xét ΔADB có 

AM/AB=AN/AD=1/2

=>MN//DB và MN=1/2BD

Xét ΔCDB có CQ/CB=CP/CD=1/2

nên PQ//BD và PQ=1/2BD

=>MN//PQ và MN=PQ=8cm

Xét ΔBAC co BM/BA=BQ/BC=1/2

=>MQ//AC và MQ=1/2AC=5cm

=>MQ vuông góc BD

=>MQ vuông góc MN

Xét tứ giác MNPQ có

MN//PQ

MN=PQ

MQ vuông góc MN

=>MNPQ là hình chữ nhật

=>\(S_{MNPQ}=5\cdot8=40\left(cm^2\right)\)

Lời giải:

$9x^2-1+(3x-1)(x+2)=0$

$\Leftrightarrow (3x-1)(3x+1)+(3x-1)(x+2)=0$

$\Leftrightarrow (3x-1)(3x+1+x+2)=0$

$\Leftrightarrow (3x-1)(4x+3)=0$

$\Leftrightarrow 3x-1=0$ hoặc $4x+3=0$

$\Leftrightarrow x=\frac{1}{3}$ hoặc $x=\frac{-3}{4}$

9x² - 1 + (3x - 1)(x + 2) = 0

\(\Leftrightarrow\) 9x² - 1 + 3x² + 6x - x - 2 = 0

\(\Leftrightarrow\) 12x² + 5x - 3 = 0

\(\Leftrightarrow\) 12x² - 4x + 9 x - 3 = 0

\(\Leftrightarrow\) (12x² - 4x) + (9x - 3) = 0

\(\Leftrightarrow\) 4x(3x - 1) + 3(3x - 1) = 0

\(\Leftrightarrow\) (3x - 1)(4x + 3) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}3x-1=0\\4x+3=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{-3}{4}\end{matrix}\right.\) 

S = \(\left\{\dfrac{1}{3},\dfrac{-3}{4}\right\}\)

Cậu tách ra `2->3` câu thôi nhe

a: =>17x-5x-15-2x-5=0

=>10x-20=0

=>x=2

b: =>\(\dfrac{3x-6-5x-10}{\left(x+2\right)\left(x-2\right)}=\dfrac{11x+23}{\left(x+2\right)\left(x-2\right)}\)

=>11x+23=-2x-16

=>13x=-39

=>x=-3(nhận)

c: =>5x+7>=3x-3

=>2x>=-10

=>x>=-5

d: =>5(3x-1)=-2(x+1)

=>15x-5=-2x-2

=>17x=3

=>x=3/17

e: =>4x^2-1-4x^2-3x-2=0

=>-3x-3=0

=>x=-1

g: =>7x-5-8x+2-7<0

=>-x-10<0

=>x+10>0

=>x>-10

\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}=-4\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+4=0\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+1+1+1+1=0\\ \Rightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2005}+\dfrac{x+2010}{2006}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2005}+\dfrac{1}{2006}\right)=0\)

mà `1/2008+1/2007+1/2005+1/2006≠ 0`

`=> x+2010=0`

`=>x=-2010`

\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\)

=>x+2010=0

=>x=-2010

`x^2=2`

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

\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{x-6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow32=\left(x-2\right)\left(x-6\right)\)

\(\Leftrightarrow32=x^2-8x+12\)

\(\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)

\(\dfrac{3x}{2}+\dfrac{11}{x}=\dfrac{4}{3}\)\(\left(dkxd:x\ne0\right)\)

\(\Leftrightarrow\dfrac{3x}{2}+\dfrac{11}{x}-\dfrac{4}{3}=0\)

\(\Leftrightarrow\dfrac{3x.3x+11.6-4.2x}{6x}=0\)

\(\Leftrightarrow9x^2+66-8x=0\)

\(\Leftrightarrow9x^2-8x+66=0\)

\(\Leftrightarrow\) Pt vô nghiệm

Vận tốc `B->A` là bao nhiêu ạ ?