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8 tháng 9 2023
d) \(2x^2+5x-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) \(\left(a+b+c=1\right)\)
LB
9 tháng 4 2020
Ta có:
\(\left(2x-1\right)^2+\left(x-2\right)\left(x+2\right)=\left(5x+1\right)\left(x-4\right)-12\)
\(\left(4x^2-4x+1\right)+x^2-4=5x^2-19x-4-12\)
\(5x^2-4x-3=5x^2-19x-16\)
\(\left(5x^2-5x^2\right)+\left(19x-4x\right)+\left(16-3\right)=0\)\(15x+13=0\)\(x=-\frac{13}{15}\)
ND
1
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 2 2020
\(\Leftrightarrow\left(x^4+x^3-2x^2\right)+\left(x^3+x^2-2x\right)+\left(6x^2+6x-12\right)=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)+x\left(x^2+x-2\right)+6\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
XO
4 tháng 7 2021
1) x2 -7x + 10 = x2 - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)
2) x2 + 3x + 2 = x2 + 2x + x + 2 = x(x + 2) + (x + 2) = (x + 1)(x + 2)
3) x2 - 7x + 12 = x2 - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)
4) x2 + 7x + 12 = x2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)
5) 16x - 5x2 - 3 = 15x - 5x2 + x - 3 = -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x)
6) 6x2 + 7x - 3 = 6x2 - 2x + 9x - 3 = 2x(3x - 1) + 3(3x - 1) = (2x + 3)(3x - 1)
7) 3x2 - 3x - 6 = 3x2 - 6x + 3x - 6 = 3x(x - 2) + 3(x - 2) = (x - 2)(3x + 3) = 3(x - 2)(x + 1)
8) 3x2 + 3x - 6 = 3x2 - 3x + 6x - 6 = 3x(x - 1) + 6(x - 1) = (x - 1)(3x + 6) = 3(x - 1)(x + 2)
9) 6x2 - 13x + 6 = 6x2 - 9x - 4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3)
10) 6x2 + 15x + 6 = 6x2 + 12x + 3x + 6 = 6x(x + 2) + 3(x + 2) = (x + 2)(6x + 3) = 3(x + 2)(3x + 1)
11) 6x2 - 20x + 6 = 6x2 - 18x - 2x + 6 = 6x(x -3) - 2(x - 3) = (6x - 2)(x - 3) = 2(3x - 1)(x - 3)
12) 8x2 + 5x - 3 = 8x2 + 8x - 3x - 3 = 8x(x + 1) - 3(x + 1) = (x + 1)(8x - 3)
TT
1
17 tháng 4 2020
P/T được xác định khi:
(5x-12+4x)-(14-x)\(\ne\)0
<=> 10x-26 \(\ne\)0
<=> x \(\ne\)2.6
PN
17 tháng 8 2020
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
NM
1
1 tháng 4 2020
b) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
<=> \(\frac{13\left(x+1\right)}{12}-\frac{5x+3}{6}=\frac{x+7}{12}\)
<=> 13(x + 1) - 2(5x + 3) = x + 7
<=> 13x + 13 - 10x - 6 = x + 7
<=> 3x + 7 = x + 7
<=> 3x + 7 - x = 7
<=> 2x + 7 = 7
<=> 2x = 7 - 7
<=> 2x = 0
<=> x = 0
c) 2x + 4(x - 2) = 5
<=> 2x + 4x - 8 = 5
<=> 6x - 8 = 5
<=> 6x = 5 + 8
<=> 6x = 13
<=> x = 13/6
11 tháng 8 2018
Bài 4 : \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
Đặt \(x^2+5x=a\) . Phương trình trở thành :
\(a^2-2a-24=0\)
\(\Leftrightarrow\left(a+4\right)\left(a-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+4=0\\a-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=-4\\a=6\end{matrix}\right.\)
Với \(a=-4\)
\(\Leftrightarrow x^2+5x=-4\)
\(\Leftrightarrow x^2+5x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)
Với \(a=6\)
\(\Leftrightarrow x^2+5x=6\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{-1;2;-3;-4\right\}\)
11 tháng 8 2018
1) x4 - 5x2 + 4 = 0
⇔ x4 - x2 - 4x2 + 4 = 0
⇔ x2(x2 - 1) - 4(x2 - 1) = 0
⇔ (x2 - 1)(x2 - 4) = 0
⇔ \(\left\{{}\begin{matrix}x^2-1=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm1\\x=\pm2\end{matrix}\right.\)
Vậy \(x=\pm1\)và \(x=\pm2\)
14 tháng 3 2017
Bài 2
Ta có :
\(x^2+5x+6=\left(x+2\right)\left(x+3\right)\)
\(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
\(x^2+9x+20=\left(x+4\right)\left(x+5\right)\)
Khi đó:
\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}=\dfrac{3}{40}\)
=> \(\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{3}{40}\)
=> \(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{3}{40}\)
=> \(\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{3}{40}\)
Giải phương trình ta được x = 3
\(=x^2-2.\frac{5}{2}x+\frac{25}{4}+\frac{23}{4}\)
\(=\left(x-\frac{5}{2}\right)^2+\frac{23}{4}\)