Tìm các giá trị của x sao cho:
a) |2x – 3| = |1 – x|
b) x2 – 4x ≤ 5
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a: |2x-3|=|1-x|
=>\(\left[{}\begin{matrix}2x-3=1-x\\2x-3=x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x+x=3+1\\2x-x=-1+3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=4\\x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
b: \(x^2-4x< =5\)
=>\(x^2-4x-5< =0\)
=>\(x^2-5x+x-5< =0\)
=>\(x\left(x-5\right)+\left(x-5\right)< =0\)
=>\(\left(x-5\right)\left(x+1\right)< =0\)
TH1: \(\left\{{}\begin{matrix}x-5>=0\\x+1< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=5\\x< =-1\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-5< =0\\x+1>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =5\\x>=-1\end{matrix}\right.\)
=>-1<=x<=5
c: 2x(2x-1)<=2x-1
=>\(\left(2x-1\right)\cdot2x-\left(2x-1\right)< =0\)
=>\(\left(2x-1\right)^2< =0\)
mà \(\left(2x-1\right)^2>=0\forall x\)
nên \(\left(2x-1\right)^2=0\)
=>2x-1=0
=>2x=1
=>\(x=\dfrac{1}{2}\)
Bài 3:
a) Ta có: \(A=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)
d) Ta có: \(D=x^2-2x+2\)
\(=x^2-2x+1+1\)
\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)
Bài 1:
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
a) \(\left|2x-3\right|=\left|1-x\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=1-x\\2x-3=x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-2\end{cases}}\)
b) \(x^2-4x\le5\)
\(\Leftrightarrow x^2-4x-5\le0\)
\(\Leftrightarrow x^2-5x+x-5\le0\)
\(\Leftrightarrow x\left(x-5\right)+\left(x-5\right)\le0\)
\(\Leftrightarrow\left(x+1\right)\left(x-5\right)\le0\)
Đến đây dễ r
c) \(2x\left(2x-1\right)\le2x-1\)
\(\Leftrightarrow2x\left(2x-1\right)-\left(2x-1\right)\le0\)
\(\Leftrightarrow\left(2x-1\right)^2\le0\)
Mà \(\left(2x-1\right)^2\ge0\)nên 2x - 1=0
Bài 2 :
a, \(x^2-4x+4+1=\left(x-2\right)^2+1\ge1\)
Dấu ''='' xảy ra khi x = 2
b, Ta có \(\left(x+1\right)^2+10\ge10\Rightarrow\dfrac{-100}{\left(x+1\right)^2+10}\ge-\dfrac{100}{10}=-10\)
Dấu ''='' xảy ra khi x = -1
Bài 1 :
a, Ta có \(A\left(x\right)=x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
b, \(B\left(x\right)=x^2\left(2x+1\right)+\left(2x+1\right)=\left(x^2+1>0\right)\left(2x+1\right)=0\Leftrightarrow x=-\dfrac{1}{2}\)
c, \(C\left(x\right)=\left|2x-3\right|=\dfrac{1}{3}\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}+3=\dfrac{10}{3}\\2x=-\dfrac{1}{3}+3=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow-4+k=-3\Leftrightarrow k=1\\ b,\Leftrightarrow-3\left(2k-18\right)=40\\ \Leftrightarrow2k-18=-\dfrac{40}{3}\Leftrightarrow k=\dfrac{7}{3}\\ c,\Leftrightarrow10+18=9\left(2+k\right)\\ \Leftrightarrow k+2=\dfrac{28}{9}\Leftrightarrow k=\dfrac{10}{9}\)
a ) \(\left|2x-3\right|=\left|1-x\right|\)
\(\Leftrightarrow2x-3=1-x\)
\(\Leftrightarrow2x+x=1+3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b ) \(x^2-4x\le5\)
\(\Rightarrow x^2-4x-5\le0\)
\(\Rightarrow\left(x-5\right)\left(x+1\right)\le0\)
+ ) \(\begin{cases}x-5>0\\x+1< 0\end{cases}\) \(\Rightarrow\begin{cases}x>5\\x< -1\end{cases}\)
+ ) \(\begin{cases}x-5< 0\\x+1>0\end{cases}\Rightarrow\begin{cases}x< 5\\x>-1\end{cases}\)
Câu a giải thiếu mak sao mọi người vẫn cho đúng ?????