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\(\begin{array}{l}a)\,{x^2} - 6x + 9 - {y^2} \\= \left( {{x^2} - 6x + 9} \right) - {y^2} \\= {\left( {x - 3} \right)^2} - {y^2} \\= \left( {x - 3 + y} \right)\left( {x - 3 - y} \right);\\b)\,4{x^2} - {y^2} + 4y - 4 = {\left( {2x} \right)^2} - \left( {{y^2} - 4y + 4} \right) \\= {\left( {2x} \right)^2} - {\left( {y - 2} \right)^2} \\= \left( {2x - y + 2} \right)\left( {2x + y - 2} \right);\\c)\,xy + {z^2} + xz + yz \\= \left( {xy + xz} \right) + \left( {{z^2} + yz} \right) \\= x\left( {y + z} \right) + z\left( {z + y} \right) \\= \left( {y + z} \right)\left( {x + z} \right);\\d)\,{x^2} - 4xy + 4{y^2} + xz - 2yz \\= \left( {{x^2} - 4xy + 4{y^2}} \right) + \left( {xz - 2yz} \right) \\= {\left( {x - 2y} \right)^2} + z\left( {x - 2y} \right) \\= \left( {x - 2y} \right)\left( {x - 2y + z} \right).\end{array}\)
B3) a) x(x-5)-4(x-5)=0
<=> (x-4)(x-5)=0
TH1 :x-4=0
<=.x=4
TH2 : x-5=0
<=>x=5
b) x(x-6)-7x-42=0
<=>x(x+6)-7(x+6)=0
<=>(x-7)(x+6)=0
th1;x-7=0
<=>x=7
th2; x+6=0
<=>x=-6
c)x^3-5x^2+x-5=0
<=> x(x^2+1)-5(x^2+1)=0
<=> (x-5)(x^2+1)=0
th1:x-5=0
<=>x=5
TH2 : x^2+1=0
<=> x^2=-1 ( vo li )
=> th2 ko tồn tại
nho thick nha
Bài 3
a, x(x-5)-4(x-5)=0
(x-4)(x-5)=0
=>\(\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
b,x(x+6)-7(x+6)=0
(x-7)(x+6)=0\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
c,x^2(x-5)+(x-5)=0
(x^2+1)(x-5)=0
\(\Rightarrow\orbr{\begin{cases}x^2+1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\in\Phi\\x=5\end{cases}}\)
a. Ta có: x2+y2-2x+4y+5=0
⇌(x-1)2+(y-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
b. Ta có: 4x2+y2-4x-6y+10=0
⇌ (2x-1)2+(y-3)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)
c.Ta có: 5x2-4xy+y2-4x+4=0
⇌(2x-y)2+(x-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=2\end{matrix}\right.\)
d.Ta có: 2x2-4xy+4y2-10x+25=0
⇌ (x-2y)2+(x-5)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{2}\\x=5\end{matrix}\right.\)
\(\text{Tìm x:}\)
\(a.x\left(x-1\right)-3x+3x=0\)
\(x\left(x-1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
\(b.3x\left(x-2\right)+10-5x=0\)
\(3x^2-6x+10-5x=0\)
\(3x^2-11x+10=0\)
\(3x^2-11x=-10\)(bn xem lại đề nhé)
\(c.x^3-5x^2+x-5=0\)
\(x^3-5x^2+x=5\)
\(d.x^4-2x^3+10x^2-20x=0\)
bài 1:phân tích thành phân tử
a> x^2-6x-y^2+9
= (x-3)^2 -y^2
= (x-3 -y) (x-3+y)
b>x^2-xy-8x+8y
= x(x-y) - 8(x-y)
= (x-8) (x-y)
c>25-4x^2-4xy-y^2
= 5^2 - (2x + y)^2
= (5 - 2x -y) (5 +2x+y)
d>xy-xz-y+z
= x(y-z) - (y-z)
= (x-1) (y-z)
e>x^2-xz-yz+2xy+y^2
= (x+y)^2 - z(x+y)
= (x+y-z) (x+y)
g>x^2-4xy+4y^2-z^2-4zt-4t^2
= (x-2y)^2 - (z + 2t)^2
= (x-2y -x-2t) (x-2y + z +2t)
bài 2:tìm X bt
a>x.(x-1)-3x+3x=0
x (x-1) =0
\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy x=0 và x=1
b>3x.(x-2)+10-5x=0
3x(x-2) - 5 (x-2)=0
(3x-5) (x-2) =0
\(\Rightarrow\hept{\begin{cases}3x-5=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}3x=5\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)
c>x^3-5x^2+x-5=0
x^2 (x-5) + (x-5) =0
(x^2 +1)(x-5) =0
\(\Rightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x^2=-1\\x=5\end{cases}\Rightarrow}\hept{\begin{cases}x\in\varphi\\x=5\end{cases}}}\)
Vậy x=5
d>x^4-2x^3+10x^2-20x=0
x^3 (x-2) + 10x(x-2) =0
(x^3 + 10x) (x-2) =0
x(x^2 + 10) (x-2) =0
\(\Rightarrow\hept{\begin{cases}x=0\\x^2+10=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-10\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x\in\varphi\\x=2\end{cases}}}}\)
Vậy x=0 và x=2
a) \(x^3-4x=0\)
\(x\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}}\)
b) \(5x\left(3x-2\right)=4-9x^2\)
\(5x\left(3x-2\right)-\left(4-9x^2\right)=0\)
\(5x\left(3x-2\right)-\left(2-3x\right)\left(2+3x\right)=0\)
\(5x\left(3x-2\right)+\left(3x-2\right)\left(2+3x\right)=0\)
\(\left(3x-2\right)\left(5x+3x+2\right)=0\)
\(\left(3x-2\right)\left(8x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-2=0\\8x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{-1}{4}\end{cases}}}\)
c) \(x^2+7x=8\)
\(x^2+7x-8=0\)
\(x^2+8x-x-8=0\)
\(x\left(x+8\right)-\left(x+8\right)=0\)
\(\left(x+8\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+8=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-8\\x=1\end{cases}}}\)
d) \(2x^2+4y^2+10x+4xy=-25\)
\(x^2+x^2+4y^2+10x+4xy+25=0\)
\(\left(4y^2+4xy+x^2\right)+\left(x^2+10x+25\right)=0\)
\(\left(2y+x\right)^2+\left(x+5\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}2y+x=0\\x+5=0\end{cases}\Rightarrow\hept{\begin{cases}y=\frac{5}{2}\\x=-5\end{cases}}}\)
a, =[(x^2)^2+2x^2+1]-x^2
=(x^2+1)^2 - x^2
=(x^2+1-x^2)(x^2+1+2x^2)
=2x^2
d,4xy+3z-12y-xz
=(4xy-12y)+(3z-xz)
=4y(x-3)-z(x-3)
=(4y-z)(x-3)