Ôn tập chương Biểu thức đại số

a: (x-3)(4-x)>0

=>(x-3)(x-4)<0

=>3<x<4

c: =>(x-3)(x-4)<0

=>3<x<4

d: \(\Leftrightarrow3x^2+3x+5x+5>0\)

\(\Leftrightarrow\left(x+1\right)\left(3x+5\right)>0\)

=>x<-5/3 hoặc x>-1

a,

f(x) có 1 nghiệm -1

=> f(x) = m.(-1)² + 5.(-1) - 2 = 0

=> m - 5 - 2 = 0

=> m = 7

b,

f(x) có nghiệm là -1

=> f(x) = m.(-1)³ + (-1)² + (-1) + 1 = 0

=> -m + 1 - 1 + 1 = 0

=> -m + 1 = 0

=> -m = -1 <=> m = 1

c,

f(x) có nghiệm là 1

=> f(x) = 1 + m² + m + m - 1 = 0

=> m² + 2m = 0

=> m(m + 2) = 0

\(\Rightarrow\left[{}\begin{matrix}m=0\\m+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-2\end{matrix}\right.\)

d,

f(x) có nghiệm là -3

=> f(x) = (-3)² - 2.(-3)² - m = 0

=> 9 - 18 - m = 0

=> -9 = m

=> m = -9

a/ \(\dfrac{x+1}{2}=\dfrac{2x+3}{5}\)

\(\Leftrightarrow5\left(x+1\right)=2\left(2x+3\right)\)

\(\Leftrightarrow5x+5=4x+6\)

\(\Leftrightarrow5x-4x=6-5\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy ...

b/ \(\left|x-1\right|+3\left|y+1\right|+\left|z+2\right|=0\)

Mà với \(\forall x;y;z\) ta có :

\(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\3\left|y+1\right|\ge0\\\left|z+2\right|\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\3\left|y+1\right|=0\\\left|z+2\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\\z+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=-2\end{matrix}\right.\)

Vậy ...

c/ \(\dfrac{x-2}{4}=\dfrac{5-3x}{4}\)

\(\Leftrightarrow x-2=5-3x\)

\(\Rightarrow x+3x=5+2\)

\(\Leftrightarrow4x=7\)

\(\Leftrightarrow x=\dfrac{7}{4}\)

Vậy ......

d/ \(\dfrac{x+2}{4}=\dfrac{4}{x+2}\)

\(\Leftrightarrow\left(x+2\right)\left(x+2\right)=16\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy ...

e/ \(\dfrac{x-1}{5}=\dfrac{-20}{x-1}\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=-100\)

\(\Leftrightarrow\left(x-1\right)^2=-100\)

Lại có : \(\left(x-1\right)^2\ge0\)

\(\Leftrightarrow\) k tồn tại x

\(\left|2x-27\right|^{2017}\ge0\\ \left(3y+10\right)^{2018}\ge0\\ \left|2x-27\right|^{2017}+\left(3y+10\right)^{2018}=0\\ \Rightarrow\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.\)

\(3^{-1}.3^x+9.3^x=28\\ \Leftrightarrow3^x\left(\dfrac{1}{3}+9\right)=28\\ \Leftrightarrow3^x.\dfrac{28}{3}=28\\ \Leftrightarrow3^x=3\\ \Leftrightarrow x=1\)

\(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)

\(\Rightarrow\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)

\(\Rightarrow\dfrac{1}{x}=\dfrac{1+2y}{6}\)

\(\Rightarrow x\left(1+2y\right)=6\)

Xét ước là xog :V

1) (1+\( \dfrac{2}{3}\)-\(\dfrac{1}{4}\)). 0,8-\(\dfrac{3}{4}\))²

=( \(\dfrac{17}{12}\).0,8-\(\dfrac{3}{4}\))²

=(\(\dfrac{17}{15}\)-\(\dfrac{3}{4}\))²

=( \(\dfrac{23}{60}\))²

^{= \(\dfrac{529}{3600}\)}

2) \(\dfrac{5}{2}\).\(\dfrac{6}{5}\)-17

=3-17=(-14)

3) (\(24\dfrac{1}{7}\)-\(33\dfrac{1}{7}\)):\(\dfrac{-3}{5}\)

=-9:(\(\dfrac{-3}{5}\))=15