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\(A=x^2-2x+10\)
\(A=\left(x^2-2x+1\right)+9\)
\(A=\left(x-1\right)^2+9\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow A\ge9\)
Dấu "=" xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy Min A = 9 khi x = 1
\(B=x^2-5x-7\)
\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow B\ge-\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)
a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0
\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)
\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)
\(-5x-8=0\)
\(x=-\frac{8}{5}\)
Mai mik làm mấy bài kia sau
Bài 3:
a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)
b: \(x^2+2x+1-y^2=\left(x+1+y\right)\left(x+1-y\right)\)
c: \(=\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
bài 1:
a) (x+1)^2-(x-1)^2-3(x+1)(x-1)
=(x+1+x-1)(x+1-x+1)-3x^2-3
=2x^2-3x^2-3
=-x^2-3
Bài 2: Tính giá trị của biểu thức sau:
\(16x^2-y^2=\left(4x+y\right)\left(4x-y\right)\)
Thay \(\hept{\begin{cases}x=87\\y=13\end{cases}}\)
\(\Rightarrow\left(4.87+13\right)\left(4.87-13\right)=361.335=120935\)
Bài 4: Tìm x
a) \(9x^2+x=0\)
\(\Rightarrow x\left(9x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\9x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{9}\end{cases}}\)
b) \(27x^3+x=0\)
\(\Rightarrow x\left(27x^2+1=0\right)\)
\(\Rightarrow\orbr{\begin{cases}x=0\\27x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\27x^2=\left(-1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{-1}{27}\end{cases}}\)
Ta có: \(\frac{-1}{27}\) loại vì \(x^2\ge0\forall x\)
Vậy \(x=0\)
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
\(x^2-25x=0\)
\(\Rightarrow x\left(x-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)
vậy_
\(\left(4x-1\right)^2-9=0\)
\(\Rightarrow\left(4x-1\right)^2-3^2=0\)
\(\Rightarrow\left(4x-1+3\right)\left(4x-1-3\right)=0\)
\(\Rightarrow\left(4x+2\right)\left(4x-4\right)=0\)
\(\Rightarrow2\cdot\left(2x+1\right)\cdot4\cdot\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}}\)
vậy_
Bài 2 :
a) \(3x^2-18x+27\)
\(=3\left(x^2-6x+9\right)\)
\(=3\left(x^2-2\cdot x\cdot3+3^2\right)\)
\(=3\left(x+3\right)^2\)
b) \(xy-y^2-x+y\)
\(=y\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(y-1\right)\)
c) \(x^2-5x-6\)
\(=x^2+x-6x-6\)
\(=x\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x-6\right)\)
Bài 1:
\(x^2-5x-6=0\)
\(\Leftrightarrow x^2+x-6x-6=0\)
\(\Leftrightarrow\left(x^2+x\right)-\left(6x+6\right)=0\)
\(\Leftrightarrow x\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)
Vậy x=-1; x=6
Bài 2:
a) Ta có: \(x+y=10\Leftrightarrow y=10-x\) (1)
Từ (1) thay vào \(P=xy\) ta được:
\(P=x\left(10-x\right)\)
\(\Leftrightarrow P=10x-x^2\)
\(\Leftrightarrow P=-x^2+10x-5^2+5^2\)
\(\Leftrightarrow P=-\left(x^2-10x+5^2\right)+5^2\)
\(\Leftrightarrow P=-\left(x-5\right)^2+25\)
Vậy GTLN của P=25 khi \(x-5=0\Leftrightarrow x=5\)
b) \(P=x^2-5x\)
\(\Leftrightarrow P=x^2-2x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\left(\dfrac{5}{2}\right)^2\)
\(\Leftrightarrow P=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\)
Vậy GTNN của \(P=\dfrac{-25}{4}\) khi \(x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)