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5 tháng 12 2021
\(\Leftrightarrow9x^2-1-1+6x^2-27x+2x-9=0\)
\(\Leftrightarrow15x^2-25x-11=0\)
\(\text{Δ}=\left(-25\right)^2-4\cdot15\cdot\left(-11\right)=1285\)
=> Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{25-\sqrt{1285}}{30}\\x_2=\dfrac{25+\sqrt{1285}}{30}\end{matrix}\right.\)
9 tháng 7 2019
b, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
\(\Rightarrow x^2-9x+20-x^2+x+2=7\)
\(\Rightarrow-8x+22=7\)
\(\Rightarrow-8x=-15\)
\(\Rightarrow x=\frac{15}{8}\)
c, \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)
\(\Rightarrow3x^2-10x+8=3x^2-27x-3\)
\(\Rightarrow3x^2-10x-3x^2+27x=\left(-3\right)+\left(-8\right)\)
\(\Rightarrow17x=-11\)
\(\Rightarrow x=-\frac{11}{17}\)
d, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27+5x-x^3=6x\)
\(\Rightarrow6x=-27\)
\(\Rightarrow x=-\frac{27}{6}\)
\(\Rightarrow x=-\frac{9}{2}\)
e, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Rightarrow3x^2-2x-5-3x^2-2x+1=x-4\)
\(\Rightarrow-4=x-4\)
\(\Rightarrow x=0\)
9 tháng 7 2019
b) (x - 5)(x - 4) - (x + 1)(x - 2) = 7
<=> x2 - 9x + 20 - x2 + x + 2 - 7 = 0
<=> 8x - 15 = 0 <=> x = 15/8
c) (3x - 4)(x - 2) = 3x(x - 9) - 3
<=> 3x2 - 10x + 8 = 3x2 - 27x - 3
<=> 17x = -11 <=> x = -11/17
d) (x - 3)(x2 + 3x + 9) + x(5 - x2) = 6x
<=> x3 - 27 - x3 + 5x - 6x = 0
<=> x = -27
e) (3x - 5)(x + 1) - (3x - 1)(x + 1) = x - 4
<=> (x + 1)(3x - 5 - 3x + 1) - x + 4 = 0
<=> -4x - 4 - x + 4 = 0 <=> x = 0
26 tháng 3 2020
a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2
= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25
= 36
b) (3x^2 - y)^2
= 9x^4 - 6x^2y + y^2
c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)
= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4
= 9x^2 + 54
d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2
= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x
= x^3 - 16x^2 + 25x
e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)
= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2
= x^3 + 2x^2 - 2x - 12
f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2
= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4
= x^6 + 2x^4 + 2x^2 + 124
YP
9 tháng 7 2019
b) \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
\(\Leftrightarrow\) \(x^2-4x-5x+20-x^2+2x-x+2\)\(=7\)
\(\Leftrightarrow\) \(-8x+22=7\)
\(\Leftrightarrow\) x= \(\frac{-15}{8}\)
4 tháng 7 2017
a, \(4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)
\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy...
b, \(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)
\(\Leftrightarrow-3x^2+15x+5x-5+3x^2=4-x\)
\(\Leftrightarrow21x=9\)
\(\Leftrightarrow x=\dfrac{3}{7}\)
Vậy...
c, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
\(\Leftrightarrow x^2-9x+20-x^2+x+2=7\)
\(\Leftrightarrow-8x=-15\Leftrightarrow x=\dfrac{15}{8}\)
Vậy...
d, \(-\left(x+3\right)\left(x-4\right)+\left(x-1\right)\left(x+1\right)=10\)
\(\Leftrightarrow-x^2+x+12+x^2-1=10\)
\(\Leftrightarrow x=-1\)
Vậy...
e, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
\(\Leftrightarrow x^3-27+5x-x^3=6x\)
\(\Leftrightarrow x=-27\)
Vậy...
HL
4 tháng 7 2017
a) \(4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)
\(4x^2-20x-7x^2+28x+3x^2-12=0\)
\(8x-12=0\)
\(4\left(2x-3\right)=0\)
\(2x-3=0\Rightarrow x=\dfrac{3}{2}\)
b) \(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)
\(-3x^2+15x+5x-5+3x^2-4+x=0\)
\(21x-9=0\)
\(3\left(7x-3\right)=0\)
\(\Rightarrow7x-3=0\Rightarrow x=\dfrac{3}{7}\)
c) \(\left(x-5\right)\left(x-4\right)-\left(x-1\right)\left(x-2\right)=7\)
\(x^2-4x-5x+20-x^2+2x+x-2-7=0\)
\(-6x+11=0\Rightarrow x=\dfrac{11}{6}\)
d) \(-\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x+1\right)=10\)
\(-x^2+4x+3x-12+x^2-1-10=0\)
\(7x-23=0\)
\(x=\dfrac{23}{7}\)
e) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
\(x^3-27+5x-x^3-6x=0\)
\(-x-27=0\Rightarrow x=-27\)
\(x\left(x-4\right)+1=3x-5\)
\(\Leftrightarrow x^2-4x=3x-6\)
\(\Leftrightarrow x^2-7x+6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-6\right)=0\)
\(\Leftrightarrow x=1,x=6\)