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1 . Phân tích đa thức thành nhân tử :
a) 4x\(^2\) + 1 - y\(^2\) - 4x
b) 2x\(^2\) - y\(^2\) + 2xy - xy
2. Tìm x :
a) \(\dfrac{1}{2}\)x\(^2\) - ( 2 - 4 ) . ( \(\dfrac{1}{2}\)x + 3 ) = 12
b) ( 4x - 1 )\(^2\) = 4
c) x . ( x - 2018 ) - 5x + 2018 . 5 = 0
3 . Tính giá trị biểu thức phụ thuộc vào biến :
B = ( x + 3 )\(^2\) - ( x + 3 ) . ( 2 - 4x ) + ( 2x - 1 ) #Hỏi cộng đồng OLM #Toán lớp 8
Bài 1:
a: \(=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
b: \(=2x\left(x+y\right)-y\left(x+y\right)=\left(x+y\right)\left(2x-y\right)\)
Câu 2:
b: =>4x-1=2 hoặc 4x-1=-2
=>x=3/4 hoặc x=-1/4
c: =>(x-2018)(x-5)=0
=>x=5 hoặc x=2018
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
1.
a) \(2x\left(x-4\right)+\left(x-1\right)\left(x+2\right)=2x^2-8x+x^2+x-2=x^2-7x-2\)
b) \(\left(x-3\right)^2-\left(x-2\right)\left(x^2+2x+4\right)=x^2-6x+9-x^3+8=-x^3+x^2-6x+17\)
2.
a) \(x^2y+xy^2-3x+3y=xy\left(x+y\right)-3\left(x-y\right)=???\)
b) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)=x\left[\left(x+y\right)^2-16\right]=\)làm tiếp chắc dễ
3.
\(\frac{x^4?2x^3+4x^2+2x+3}{x^2+1}\) Giữa x^4 và 2x^3 (vị trí dấu ? là dấu + hay -)
4) \(A=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)
\(A\ge\frac{7}{4}\)
Vậy GTNN của A là 7/4
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
\(1.\)
\(x^2-2x+1-xy-y=\left(x-1\right)^2-y\left(x-1\right)=\left(x-1\right)\left(x-1-y\right)\)
\(2.\)
\(x^3-4x^2+4x-2x+2=x\left(x^2-4x+4\right)-2\left(x-1\right)=x\left(x-2\right)^2-2\left(x-1\right)\)
\(3.\)
\(10x-25-x^2+4y^2=4y^2-\left(x^2-10x+25\right)=4y^2-\left(x-5\right)^2=\left(2y+x-5\right)\left(2y-x+5\right)\)
\(4.\)
\(4x^2-2x+2xy-y=2x\left(2x-1\right)+y\left(2x-1\right)=\left(2x-1\right)\left(2x+y\right)\)
\(5.\)
\(4x\left(x-3\right)^2-3x^2+9x=4x\left(x-3\right)^2-3x\left(x-3\right)=\left(x-3\right)\left(4x^2-12x-3x\right)\)
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
\(a,x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
\(b,x\left(y+1\right)+y+1=\left(x+1\right)\left(y+1\right)\)
\(c,\left(x+y\right)^2-9z^2=\left(x+y-3z\right)\left(x+y+3z\right)\)
\(d,5\left(x^2-2x+1-y^2\right)=5\left(\left(x+1\right)^2-y^2\right)=5\left(x+1-y\right)\left(x+1+y\right)\)
1) a) ta có : \(4x^2+1-y^2-4x\Leftrightarrow\left(2x-2\right)^2-y^2=\left(2x-2-y\right)\left(2x-2+y\right)\)
b) \(2x^2-y^2+2xy-xy\Leftrightarrow2x\left(x+y\right)-y\left(x+y\right)=\left(2x-y\right)\left(x+y\right)\)
bài 2 : a) ta có : \(\dfrac{1}{2}x^2+2\left(\dfrac{1}{2}x+3\right)-12=0\Leftrightarrow\dfrac{1}{2}x^2+x-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1+\sqrt{13}\\x=-1-\sqrt{13}\end{matrix}\right.\) câu này mk nghỉ đề sai
b) ta có : \(\left(4x-1\right)^2=4\Leftrightarrow\left[{}\begin{matrix}4x-1=2\\4x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
c) ta có : \(x\left(x-2018\right)-5x+2018.5=0\Leftrightarrow x^2-2023x+10090=0\)
\(\Leftrightarrow\left(x-2018\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2018\\x=5\end{matrix}\right.\)
bài 3 câu này bn chỉ cần nhân tung ra rồi rút gọn lại ra số là kết luận đc .
Bài 1:
\(a,4x^2+1-y^2-4x\)
\(=\left(4x^2-4x+1\right)-y^2\)
\(=\left(2x-1\right)^2-y^2\)
\(=\left(2x-1-y\right)\left(2x-1+y\right)\)
\(b,2x^2-y^2+2xy-xy\)
\(=\left(2x^2+2xy\right)-\left(y^2+xy\right)\)
\(=2x\left(x+y\right)-y\left(x+y\right)\)
\(=\left(x+y\right)\left(2x-y\right)\)
Bài 2:
\(a,\dfrac{1}{2}x^2-\left(2-4\right).\left(\dfrac{1}{2}x+3\right)=12\)
\(\Leftrightarrow\dfrac{1}{2}x^2+2\left(\dfrac{1}{2}x+1\right)=12\)
\(\Leftrightarrow\dfrac{1}{2}x^2+x+2=12\)
\(\Leftrightarrow\dfrac{1}{2}x^2+x-10=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{2}}x\right)^2+2.\dfrac{1}{\sqrt{2}}x.\dfrac{1}{\sqrt{2}}+\dfrac{1}{2}-\dfrac{1}{2}-10=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{2}}x+\dfrac{1}{\sqrt{2}}\right)^2-\dfrac{21}{2}=0\)
cái này vẫn có thể giải tiếp đc nhg mk thấy nếu bn hok lớp 8 thì chưa đã hok đến cái này nên mk nghĩ bn nên kt lại đề bài
\(b,\left(4x-1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=2\\4x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(c,x\left(x-2018\right)-5x+2018.5=0\)
\(\Leftrightarrow x\left(x-2018\right)-5\left(x-2018\right)=0\)
\(\Leftrightarrow\left(x-2018\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2018\\x=5\end{matrix}\right.\)
Bài 3: bn ơi đề sai