tim X Phân tích thành nhân tử
a/ x+5x2=0
b/ 5x2 -13x=0
x+1=(x+1)2
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15 tháng 10 2021
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
15 tháng 10 2021
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
5 tháng 9 2021
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
5 tháng 9 2021
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
29 tháng 7 2021
1/
a)5x – 20y=5(x-4y)
b) 5x.(x – 1) – 3x(x – 1)=2x(x-1)
c) x.(x+y) – 5x – 5y=c) x.(x+y) – 5(x+y)=(x-5)(x+y)
2/
a)x2 + xy + x = x(x+y+1)=77.(77+22+1)=77.100=7700
b) x . ( x – y ) + y . ( y – x )=(x-y)(x-y)=(x-y)2=(53-3)2=2500
3/
a) X + 5x2 = 0
⇒x(x+5)=0
⇒hoặc x=0
x+5=0⇒x=-5
b)x + 1 = ( x + 1 )2
⇒(x + 1)-( x + 1 )2 =0
⇒x(x+1)=0
⇒ hoặc x=0
hoặc x+1=0⇒x=-1
T
1
T
13 tháng 1
Bài 1:
\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)
\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)
\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)
T
13 tháng 1
Bài 2:
\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)
Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:
\(\left(y-1\right)\left(y+1\right)=120\)
\(\Leftrightarrow y^2-1=120\)
\(\Leftrightarrow y^2=121\)
\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)
+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)
+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)
\(\Leftrightarrow x^2+5x+16=0\)
\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)
Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
\(\Rightarrow\) loại
Vậy \(x\in\left\{1;-6\right\}\).
\(b,\) Đề thiếu vế phải rồi bạn.
11 tháng 11 2021
a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
8 tháng 9 2021
d: \(x\left(x^2-1\right)+3\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
e: \(x^2-10x+25=\left(x-5\right)^2\)
g: \(x^2-64=\left(x-8\right)\left(x+8\right)\)
h: \(\left(x+y\right)^2-\left(x^2-y^2\right)\)
\(=\left(x+y\right)\left(x+y-x+y\right)\)
\(=2y\left(x+y\right)\)
i: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
k: \(x^2+2xy+y^2-25=\left(x+y-5\right)\left(x+y+5\right)\)
l: \(2xy-x^2-y^2+16\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left(x-y-4\right)\left(x-y+4\right)\)
8 tháng 9 2021
a: \(5x-15y=5\left(x-3y\right)\)
b: \(5x^2y^2+15x^2y+30xy^2=5xy\left(xy+3x+6y\right)\)
c: \(x^3-2x^2y+xy^2-9x\)
\(=x\left(x^2-9-2xy+y^2\right)\)
\(=x\left(x-y-3\right)\left(x-y+3\right)\)
25 tháng 10 2021
Câu 3:
a: 2x-8=4
nên 2x=12
hay x=6
b: 7x-3x=2x+7
\(\Leftrightarrow4x-2x=7\)
hay \(x=\dfrac{7}{2}\)
Mình làm ý b,c thôi a tương tự b
b) 5x^2 - 13x = 0
=> x(5x - 1 3) = 0
=> x = 0 hoặc 5x - 13 = 0
=> x = 0 hoặc x = 13/5
b) x + 1 = ( x+1 )^2
=> (x + 1 )^2 - (x+ 1) = 0
=> (x +1 )( x + 1 - 1 ) = 0
=> x(x + 1 ) = 0
=> x= 0 hoặc x + 1 = 0
=> x = 0 hoặc x = -1
a, x+5x2=0
<=>x(1+5x)=0
<=>x=0 hoặc 1+5x=0
<=>x=0 hoặc x=-1/5
b, 5x2-13x=0
<=>x(5x-13)=0
<=>x=0 hoặc 5x-13=0
<=>x=0 hoặc x=13/5
c, x+1=(x+1)2
<=>(x+1)2-(x+1)=0
<=>(x+1)(x+1-1)=0
<=>x(x+1)=0
<=>x=0 hoặc x+1=0
<=>x=0 hoặc x=-1