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7 tháng 11 2021
e) \(\left(9x^2-49\right)+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\text{[}\left(3x\right)^2-7^2\text{]}+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\left(3x-7\right)\left(3x+7\right)+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\left(3x+7\right)\text{[}\left(3x-7\right)+\left(7x+3\right)\text{]}=0\)
\(\Rightarrow\left(3x+7\right)\left(3x-7+7x+3\right)=0\)
\(\Rightarrow\left(3x+7\right)\left(10x-4\right)=0\)
=> 2 TH
*3x+7=0 *10x-4=0
=>3x=-7 =>10x=4
=>x=-7/3 =>x=4/10=2/5
vậy x=-7/3 hoặc x=2/5
g) \(\left(x-4\right)^2=\left(2x-1\right)^2\)
\(\Rightarrow\left(x-4\right)^2-\left(2x-1\right)^2=0\)
\(\Rightarrow\left(x-4-2x+1\right)\left(x-4+2x-1\right)=0\)
\(\Rightarrow\left(-x-3\right)\left(3x-5\right)=0\)
\(\Rightarrow-\left(x+3\right)\left(3x-5\right)=0\)
=> 2 TH
*-(x+3)=0 *3x-5=0
=>-x=-3 =>3x=5
=x=3 =>x=5/3
h)\(x^2-x^2+x-1=0\)
\(\Rightarrow0+x-1=0\)
\(\Rightarrow x-1=0\)
=>x=0+1
=>x=1
vậy x=1
k, x(x+ 16) - 7x - 42 = 0
=>x^2+16x-7x-42=0
=>x^2+9x-42=0
vì x^2>0
do đó x^2+9x-42>0
nên o có gt nào của x t/m y/cầu đề bài
m)x^2+7x+12=0
=>x^2+3x++4x+12=0
=>x(x+3)+4(x+3)=0
=>(x+4).(x+3)=0
=>2 TH
=> *x+4=0
=>x=-4
vậy x=-4
*x+3=0
=>x=-3
vậy x=-3
n)x^2-7x+12=0
=>x^2-4x-3x+12=0
=>x(x-4)-3(x-4)=0
=>(x-3).(x-4)=0
=>2 TH
*x-3=0=>x=0+3=>x=3
*x-4=0=>x=0+4=>x=4
vậy x=3 hoặc x=4
KA
7 tháng 11 2021
a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1
b)(x+1)(x+2)(x+5)−x2(x+8)=27⇔x2+2x+x+2(x+5)−x3−8x2=27⇔x2(x+5)+2x(x+5)+x(x+5)+2(x+5)−x3−8x2=27⇔x3+5x2+2x2+10x+x2+5x+2x+10−x3−8x2=27⇔17x+10=27⇔17x=17⇒x=1
15 tháng 8 2019
Thực hiện phép chia ta có:
Ta có: \(x^3-2x^2+7x-7=\left(x^2+3\right)\left(x-2\right)+4x-1\)
\(x^3-2x^2+7x-7\) chia hết cho \(x^2+3\)
=> \(4x-1⋮x^2+3\) (1)
=> \(4x^2-x=x\left(4x-1\right)⋮x^2+3\)
Mà: \(4x^2+12=4\left(x^2+3\right)⋮x^2+3\)
=> \(\left(4x^2-x\right)-\left(4x^2+12\right)⋮x^2+3\)
=> \(-x-12⋮x^2+3\)
=> \(x+12⋮x^2+3\)
=> \(4x+48⋮x^2+3\) (2)
Từ (1); (2) => \(\left(4x+48\right)-\left(4x-1\right)⋮x^2+3\)
=> \(49⋮x^2+3\)
=> \(x^2+3\in\left\{\pm1;\pm7;\pm49\right\}\) vì \(x^2+3\ge3\) với mọi x
=> \(\begin{cases}x^2+3=7\\x^2+3=49\end{cases}\Rightarrow\orbr{\begin{cases}x^2=4\\x^2=46\left(loại\right)\end{cases}}\)
Với \(x^2=4\Rightarrow x=\pm2\) thử vào bài toán x=-2 loại. x=2 thỏa mãn
Vậy x=2
3 tháng 8 2017
\(2x^2-7x+3=2x^2-6x-x+3=2x\left(x-3\right)-\left(x-3\right)=\left(2x-1\right)\left(x-3\right)\)
\(3x^2+7x-76=3x^2-12x+19x-76=3x\left(x-4\right)+19\left(x-4\right)=\left(3x+19\right)\left(x-4\right)\)
\(\dfrac{1}{2}x^2-\dfrac{19}{6}x+1=\dfrac{1}{2}x^2-3x-\dfrac{1}{6}x+1=x\left(\dfrac{x}{2}-3\right)-\dfrac{1}{3}\left(\dfrac{x}{2}-3\right)=\left(x-\dfrac{1}{3}\right)\left(\dfrac{x}{2}-3\right)\)
\(2x^2-5x-3\): sai đề
\(15x^2-x-6=15x^2-10x+9x-6=5x\left(3x-2\right)+3\left(3x-2\right)=\left(5x+3\right)\left(3x-2\right)\)
\(3x^2+5x-2=3x^2+6x-x-2=3x\left(x+2\right)-\left(x+2\right)=\left(3x-1\right)\left(x+2\right)\)
DT
27 tháng 8 2020
a) (7x + 4)2 - (7x + 4)(7x - 4)
= 49x2 + 56x + 16 - 49x2 + 16
= 56x + 32
b) (x - 2y)3 - 6xy(x - 2y)
= x3 - 6x2y + 12xy2 - 8y3 - 6x2y + 12xy2
= x3 - 12x2y + 24xy2 - 8y3
c) (3x + y)(9x2 - 3xy + y2) - (3xy)3 - 27x2y
= 27x3 + y3 - (3xy)3 - 27x2y
d) 5(x + 3)(x - 3) + (2x + 3)2 + (x - 6)2
= 5x2 - 45 + 4x2 + 12x + 9 + x2 - 12x + 36
= 10x2
e) (2x + 3)2 + (2x - 3)2 - 2(4x2 - 9)
= (2x + 3)2 + (2x - 3)2 - 2(2x - 3)(2x + 3)
= (2x + 3 - 2x + 3)2
= 62 = 36
g) (x + 2)3 + (x - 2)3 + x3 - 3x(x - 2)(x + 2)
= (x+2+x-2)(x2 + 4x + 4 - x2 + 4 + x2 - 4x + 4) + x3 - 3x3 + 12x
= 2x(x2 + 8) + x3 - 3x3 + 12x
= 2x3 + 16x + x3 - 3x3 + 12x
= 28x
T
17 tháng 7 2019
\(x^2-6x+5=0\)
<=> \(x^2-x-5x+5=0\)
<=> \(x\left(x-1\right)-5\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x-5\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy phương trình có nghiệm là x=1 và x=5
\(2x^2+7x-9=0\) ( nếu là 9 thì ko ra kq đc nên mình đổi thành -9 nha )
<=> \(2x^2-2x+9x-9=0\)
<=> \(2x\left(x-1\right)+9\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(2x+9\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\2x+9=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=\frac{-9}{2}\end{matrix}\right.\)
\(4x^2-7x+3=0\)
<=> \(4x^2-4x-3x+3=0\)
<=>\(4x\left(x-1\right)-3\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(4x-3\right)=0\)
<=> \(\left\{{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
\(2\left(x+5\right)=x^2+5x\)
<=> \(2\left(x+5\right)-x^2-5x=0\)
<=>\(2\left(x+5\right)-x\left(x+5\right)=0\)
<=>\(\left(x+5\right)\left(2-x\right)=0\)
<=>\(\left\{{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
BH
20 tháng 2 2019
\(\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1}{\left(x-3\right)\left(2x-1\right)}=\frac{2x+5}{\left(x-3\right)\left(2x-1\right)}\)
\(\frac{\left(x-3\right)\left(x+4\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=\frac{\left(2x+5\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}\)
\(\Rightarrow x^2+x-12+x^2-x-2=2x^2+x-10\Leftrightarrow x=-4\)
20 tháng 2 2019
\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{2x-5}{2x^2-7x+3}-\frac{x+1}{2x^2-7x+3}\)
\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{x+4}{2x^2-7x+3}\)
TH1:\(x+4\ne0\)
\(\Rightarrow2x^2-5x+2=2x^2-7x+3\)
\(\Rightarrow-5x+2=-7x+3\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
TH2:\(x+4=0\)
\(\Rightarrow x=-4\)
2 tháng 6 2022
Câu 1:
\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)
6 tháng 6 2019
/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0
⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0
⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0
Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:
(t-18).t +17=0
⇔t2−18t+17=0⇔t2−18t+17=0
⇔(t−17)(t−1)=0⇔(t−17)(t−1)=0
⇔[t=17(tm)t=1(tm)⇔[t=17(tm)t=1(tm) ⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0
⇔[(20x+9−341−−−√)(20x+9+341−−−√)=0(20x+9−21−−√)(20x+9+21−−√)=0⇔[(20x+9−341)(20x+9+341)=0(20x+9−21)(20x+9+21)=0
⇔⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x=−9+341−−−√20x=−9−341−−−√20x=−9+21−−√20x=−9−21−−√20
6 tháng 6 2019
\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)
\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)
\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)
Đặt ....
\(=\left(2x^3-4x^2-3x^2+6x+x-2\right):\left(x-2\right)\\ =\left(x-2\right)\left(2x^2-3x+1\right):\left(x-2\right)=2x^2-3x+1\)