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GN
7
N
22 tháng 10 2018
\(x^3-11x^2+30x=0\)
\(\left(x-6\right).\left(x-5\right).x=0\)
\(=>\orbr{\begin{cases}x-6=0\\x-5=0,x=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=6\\x=5,x=0\end{cases}}\)
P/S: mk mới lớp 7 sai sót mong bỏ qua
N
22 tháng 10 2018
\(8x^2+30x+7=0\)
\(8x^2+28x+2x+7=0\)
\(2x.\left(4x+1\right)+7.\left(4x+1\right)=0\)
\(\left(2x+7\right).\left(4x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=-7\\4x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)
vậy ....
P/S sorry mk làm hơi lâu :)__chờ tí làm câu a cho
BC
1
16 tháng 10 2018
b) 8x2 + 30x + 7 = 0
8x2 + 16x + 14x + 7 = 0
8x.(x+2) + 7.(x+2) = 0
(x+2).(8x+7) = 0
..
bn tự làm tiếp nhé! ^-^
c) x3 - 11x2 + 30x = 0
x.(x2 - 11x +30) = 0
\(x.\left(x^2-5x-6x+30\right)=0.\)
x.[ x.(x-5) - 6.(x-5) ] = 0
x.(x-5).(x-6) = 0
...
30 tháng 9 2018
a) \(2x^2+5x-18\)
\(=2x^2-4x+9x-18\)
\(=2x\left(x-2\right)+9\left(x-2\right)\)
\(=\left(x-2\right)\left(2x+9\right)\)
b) \(4x^2-17x+15\)
\(=4x^2-12x-5x+15\)
\(=4x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(4x-5\right)\)
c) \(-8x^2+10x+7\)
\(=-8x^2-4x+14x+7\)
\(=-4x\left(2x+1\right)+7\left(2x+1\right)\)
\(=\left(2x+1\right)\left(-4x+7\right)\)
d) \(7x^2-30x+8\)
\(=7x^2-28x-2x+8\)
\(=7x\left(x-4\right)-2\left(x-4\right)\)
\(=\left(x-4\right)\left(7x-2\right)\)
e) \(-x^3+11x^2-30x\)
\(=x\left(-x^2+11x-30\right)\)
\(=x\left(-x^2+5x+6x-30\right)\)
\(=x\left[-x\left(x-5\right)+6\left(x-5\right)\right]\)
\(=x\left(x-5\right)\left(-x+6\right)\)
30 tháng 9 2018
a) 2x\(^2\) + 5x - 18 = 2x\(^2\) + 9x - 4x - 18 = x(2x + 9) - 2(2x + 9) = (x-2)(2x-9)
b) 4x\(^2\) - 17x - 15 = 4x\(^2\) + 20x - 3x - 15 = 4x(x + 5 ) - 3(x + 5) = (4x - 3 )(x + 5)
c) -8x\(^2\) + 10x + 7 = -8x\(^2\) + 14x - 4x + 7 =-2x(4x - 7) - (4x - 7) = (-2x - 1)(4x - 7)
d) 7x\(^2\) - 30x + 8 = 7x\(^2\) + 2x + 28x + 8 = x(7x + 2) + 4(7x + 2) = (x + 4)(7x + 2)
e) - x\(^3\) + 11x\(^2\) - 30x = -x(x\(^2\) - 11x + 30) = -x(x\(^2\) - 5x - 6x + 30) = -x\(\left[x\left(x-5\right)-6\left(x-5\right)\right]\) = -x(x-6)(x-5)
NA
13 tháng 6 2017
a)
a) 3x2+12x−66=0
=> 3(x + 2)2 - 12 - 66 = 0
=> 3(x + 2)2 - 78 = 0
=> 3(x + 2)2 = 78
=> (x + 2)2 = 26
=> x = \(\sqrt{26}-2\)
b)9x2−30x+225=0
=> (3x - 5)2 - 25 + 225 = 0
=> (3x - 5)2 + 200 = 0
=> (3x - 5)2 = -200
9x2 - 30x + 225 không có ngiệmc)x2+3x−10=0=> (x + 1,5)2 - 2,25 - 10 = 0
=> (x + 1,5)2 - 12,25 = 0
=> (x + 1,5)2 = 12, 25
=> x + 1,5 = 3,5
=> x = 2
d)3x2−7x+1=0=> 3(x - \(\dfrac{7}{6}\))2 - \(\dfrac{49}{12}\) + 1 = 0
=> 3(x - \(\dfrac{7}{6}\))2 - \(\dfrac{37}{12}\) = 0
=> 3(x - \(\dfrac{7}{6}\))2 = \(\dfrac{37}{12}\)
=> (x - \(\dfrac{7}{6}\))2 = \(\dfrac{37}{36}\)
=> x = \(\dfrac{\sqrt{37}}{6}+\dfrac{7}{6}=\dfrac{\sqrt{37}+7}{6}\)
e) 3x2−7x+8=0
=> 3(x - \(\dfrac{7}{6}\))2 - \(\dfrac{49}{12}\)+ 8 = 0
=> 3(x - \(\dfrac{7}{6}\))2 + \(\dfrac{47}{12}\) = 0
=> 3(x - \(\dfrac{7}{6}\))2 = \(-\dfrac{47}{12}\)
KL : Không có ngiệm
TN
12 tháng 3 2020
\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\\\Leftrightarrow\left(x+1\right)^2=4\left(x-1\right)^2\\\Leftrightarrow \left(x+1\right)^2-4\left(x-1\right)^2=0\\\Leftrightarrow \left(x+1\right)^2-\left(2x-2\right)^2=0\\\Leftrightarrow \left[\left(x+1\right)+\left(2x-2\right)\right]\left[\left(x+1\right)-\left(2x-2\right)\right] =0\\ \Leftrightarrow\left(x+1+2x-2\right)\left(x+1-2x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(3-x\right)=0\\\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{3};3\right\}\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\\ \Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\\ \Leftrightarrow\left[\left(2x+7\right)+\left(3x+6\right)\right]\left[\left(2x+7\right)-\left(3x+6\right)\right]=0\\ \Leftrightarrow\left(2x+7+3x+6\right)\left(2x+7-3x-6\right)=0\\ \Leftrightarrow\left(5x+13\right)\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+13=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-13}{5}\\x=1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-13}{5};1\right\}\)
\(4\left(2x+7\right)^2=9\left(x+3\right)^2\\\Leftrightarrow 4\left(2x+7\right)^2-9\left(x+3\right)=0\\ \Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\\\Leftrightarrow \left[\left(4x+14\right)+\left(3x+9\right)\right]\left[\left(4x+14\right)-\left(3x+9\right)\right]=0\\\Leftrightarrow \left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\\\Leftrightarrow \left(7x+23\right)\left(x+5\right)=0\\\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-23}{7};-5\right\}\)
AH
Akai Haruma
Giáo viên
15 tháng 7 2018
a)
\(x^3-7x-6=x^3-x-6x-6\)
\(=x(x^2-1)-6(x+1)\)
\(=x(x-1)(x+1)-6(x+1)=(x+1)[x(x-1)-6]\)
\(=(x+1)(x^2-x-6)=(x+1)[x^2-3x+2x-6]\)
\(=(x+1)[x(x-3)+2(x-3)]=(x+1)(x+2)(x-3)\)
b) \(x^3-6x^2+8x\)
\(=x(x^2-6x+8)\)
\(=x(x^2-4x-2x+8)\)
\(=x[x(x-4)-2(x-4)]=x(x-2)(x-4)\)
AH
Akai Haruma
Giáo viên
15 tháng 7 2018
c) \(x^4+2x^3-16x^2-2x+15\)
\(=(x^4+2x^3-x^2-2x)-15x^2+15\)
\(=[(x^4-x^2)+(2x^3-2x)]-15(x^2-1)\)
\(=[x^2(x^2-1)+2x(x^2-1)]-15(x^2-1)\)
\(=(x^2-1)(x^2+2x)-15(x^2-1)=(x^2-1)(x^2+2x-15)\)
\(=(x^2-1)(x^2-3x+5x-15)=(x^2-1)[x(x-3)+5(x-3)]\)
\(=(x^2-1)(x+5)(x-3)=(x-1)(x+1)(x+5)(x-3)\)
d)
\(x^3-11x^2+30x=x(x^2-11x+30)\)
\(=x(x^2-5x-6x+30)\)
\(=x[x(x-5)-6(x-5)]=x(x-6)(x-5)\)
AN
6 tháng 10 2018
a)\(x^2-13x+36=x^2-4x-9x+36=x\left(x-4\right)-9\left(x-4\right)=\left(x-9\right)\left(x-4\right)\)
b)\(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x+6\right)\left(x-3\right)\)
c)\(x^2-5x-24=x^2+3x-8x-24=x\left(x+3\right)-8\left(x+3\right)=\left(x-8\right)\left(x+3\right)\)
d)\(3x^2-16x+5=3x^2-x-15x+5=x\left(3x-1\right)-5\left(3x-1\right)=\left(x-5\right)\left(3x-1\right)\)
e)\(8x^2+30x+7=8x^2+28x+2x+7=4x\left(2x+7\right)+\left(2x+7\right)=\left(4x+1\right)\left(2x+7\right)\)
g)\(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(2x-1\right)\left(x-3\right)\)
h)\(6x^2-7x+3=6x^2-9x-2x+3=3x\left(2x-3\right)-\left(2x-3\right)=\left(3x-1\right)\left(2x-3\right)\)
i)\(3x^2-14x+11=3x^2-3x-11x+11=3x\left(x-1\right)-11\left(x-1\right)=\left(3x-11\right)\left(x-1\right)\)
k)\(5x^2+8x-13=5x^2-5x+13x-13=5x\left(x-1\right)+13\left(x-1\right)=\left(5x+13\right)\left(x-1\right)\)
KB
6 tháng 10 2018
a ) \(x^2-13x+36=x^2-4x-9x+36=x\left(x-4\right)-9\left(x-4\right)=\left(x-9\right)\left(x-4\right)\)
b ) \(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x+6\right)\left(x-3\right)\)
c ) \(x^2-5x-24=x^2-3x+8x-24=x\left(x-3\right)+8\left(x-3\right)=\left(x+8\right)\left(x-3\right)\)
d ) \(3x^2-16x+5=3x^2-15x-x+5=3x\left(x-5\right)-\left(x-5\right)=\left(3x-1\right)\left(x-5\right)\)
e ) \(8x^2+30x+7=8x^2+2x+28x+7=2x\left(4x+1\right)+7\left(4x+1\right)=\left(2x+7\right)\left(4x+1\right)\)
g ) \(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(2x-1\right)\left(x-3\right)\)
h ) \(6x^2-7x-20=6x^2-15x+8x-20=3x\left(2x-5\right)+4\left(2x-5\right)=\left(3x+4\right)\left(2x-5\right)\)
i ) \(3x^2-14x+11=3x^2-3x-11x+11=3x\left(x-1\right)-11\left(x-1\right)=\left(3x-11\right)\left(x-1\right)\)
k ) \(5x^2+8x-13=5x^2-5x+13x-13=5x\left(x-1\right)+13\left(x-1\right)=\left(5x+13\right)\left(x-1\right)\)
\(x^2+3x-18=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}}\)
\(8x^2+30x+7=0\)
\(\Leftrightarrow\left(x+\frac{1}{4}\right)\left(x+\frac{7}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=0\\x+\frac{7}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=-\frac{7}{2}\end{cases}}}\)
\(x^3-11x^2+30x=0\)
\(\Leftrightarrow x\left(x-6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}}\)hoặc \(x=0\)
\(x^2+3x-18=x^2-3x+6x-18=x\left(x-3\right)+6\left(x-3\right)=\left(x-2\right)\left(x+6\right)\)