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a: \(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
=>4x-27=1
hay x=7
b: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x+1\right)^2+3x^2=15\)
\(\Leftrightarrow-9x^2+27x+6x^2+12x+6+3x^2=15\)
=>39x+6=15
hay x=3/13
c: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(\Leftrightarrow3x-40=2\)
hay x=14
a) (x + 2)(x + 3) - (x - 2)(x + 5) = 6
x2 + 3x + 2x + 6 - (x2 + 5x - 2x - 10) = 6
x2 + 5x + 6 - x2 - 3x + 10 = 6
2x +16 = 6
\(\Rightarrow\) 2x = -10
\(\Rightarrow\) x = -5
b) (3x + 2)(2x + 9) - (x + 2)(6x + 1) = (x + 1) - (x - 6)
6x2 + 27x + 4x + 18 - (6x2 + x + 12x + 2) = x + 1 - x + 6
6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
18x + 16 = 7
\(\Rightarrow\) 18x = -9
\(\Rightarrow\) x = -0.5
c) 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) = 0
3(6x2 - 2x - 3x + 1) - (18x2 - 2x - 27x + 3) = 0
3(6x2 - 5x + 1) - (18x2 - 29x + 3) = 0
18x2 - 15x + 3 - 18x2 + 29x - 3 = 0
14x = 0
\(\Rightarrow\) x = 0
Câu 1 :
\(\left(2x+y\right)\left(4x^2-2xy+y^2\right)=\left(2x\right)^3+y^3=8x^3+y^3\)Câu 2:
\(A=3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)\(\Leftrightarrow3\left(6x^2-2x-6\right)-2\left(4x^2+13x-12\right)+36x-9x^2=0\)\(\Leftrightarrow18x^2-6x-18-8x^2-26x+24+36x-9x^2=0\)\(\Leftrightarrow x^2+4x+6=0\)
\(\Leftrightarrow\left(x+2\right)^2=-2\)
Ta có:
\(\left(x+2\right)^2\ge0\forall x\)
Vậy pt vô nghiệm
Vậy:ko......
Câu 3:
\(\left(5x-3\right)\left(7x+2\right)-35x\left(x-1\right)=42\)
\(\Leftrightarrow35x^2+10x-21x-6-35x^2+35x-42=0\)\(\Leftrightarrow14x=48\Leftrightarrow x=\dfrac{7}{24}\)
Câu 4:
\(\left(3x+5\right)\left(2x-1\right)+\left(5-6x\right)\left(x+2\right)=x\)
\(\Leftrightarrow6x^2-3x+10x-5+5x+10-6x^2-12x-x=0\)\(\Leftrightarrow-x=-5\Rightarrow x=5\)
câu 6,
Câu 6: \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(\Rightarrow10x^2+9x-\left(10x^2-2x+15x-3\right)=8\)
\(\Rightarrow10x^2+9x-10x^2+2x-15x+3=8\)
\(\Rightarrow-4x+3=8\)
\(\Rightarrow-4x=5\Rightarrow x=\dfrac{-5}{4}\)
Câu 7: \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)
\(\Rightarrow\left(x^2+x\right)\left(x+6\right)-x^3=5x\)
\(\Rightarrow x^3+x^2+6x^2+6x-x^3=5x\)
\(\Rightarrow7x^2=-x\)
\(\Rightarrow7x=-1\Rightarrow x=\dfrac{-1}{7}\).
1. (3x - 5)2 - (3x + 1)2 = 8
=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8
=> -6(6x - 4) = 8
=> 6x - 4 = \(\dfrac{-4}{3}\)
\(\Rightarrow x=\dfrac{4}{9}\)
2) 2x(8x - 3) - (4x - 3)2 = 27
=> 16x2 - 6x - 16x2 + 24x - 9 = 27
=> 18x - 9 = 27
=> x = 2
3) (2x - 3)2 - (2x + 1)2 = 3
=> (2x - 3 - 2x - 1)(2x - 3 + 2x +1) = 3
=> -4(4x - 2) = 3
=> 4x - 2 = \(\dfrac{-3}{4}\)
\(\Rightarrow x=\dfrac{5}{16}\)
4) (x + 5)2 - x2 = 45
=> (x + 5 - x)(x + 5 + x) = 45
=> 5(2x + 5) = 45
=> 2x + 5 = 9
=> x = 2
5) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x + 1)2 = 18
=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9(x2 + 2x + 1) = 18
=> -9x2 + 27x + 9x2 + 18x + 9 = 18
=> 45x + 9 = 18
=> 45x = 9
=> x = \(\dfrac{1}{5}\)
6) x(x - 4)(x + 4) - (x - 5)(x2 + 5x + 25) = 13
=> x (x2 - 16) - (x3 - 125) = 13
=> x3 - 16x - x3 + 125 = 13
=> -16x = -112
=> x = 7.
C =(\(x-1\))\(^3\) - (\(x-3\))(\(x^2+6x+9\)) - 3\(x\left(1-x\right)\)
C = (\(x-1\))\(^3\) - (\(x-3\)).[\(x+3\)]\(^2\) - 3\(x\)(1 - \(x\))
C = \(x^3-3x^2+3x-1\)- (\(x^2-9)\left(x+3\right)\) - 3\(x\) + 3\(x^2\)
C = \(x^3\) - 3\(x^2\) + 3\(x-1\) - \(x^3\) - 3\(x^2\) + 9\(x+27\) - 3\(x\) + 3\(x^2\)
C = (\(x^3\) - \(x^3\)) - (3\(x^2\) + 3\(x^2-3x^2\)) + (3\(x\) - 3\(x+9x\)) + (-1 + 27)
C = 0 - (6\(x^2\) - 3\(x^2\)) + (0 + 9\(x\)) + 26
= - 3\(x^2\) + 9\(x\) + 26
Ta có: \(C=\left(x-1\right)^3-\left(x-3\right)\left(x^2+6x+9\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-\left(x-3\right)\left(x+3\right)^2-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-3x+3x^2-\left(x^2-9\right)\left(x+3\right)\)
\(=x^3-1-x^3-3x^2+9x+27=-3x^2+9x+26\)
C =(\(x - 1\))\(^{3}\) - (\(x - 3\))(\(x^{2} + 6 x + 9\)) - 3\(x \left(\right. 1 - x \left.\right)\)
C = (\(x - 1\))\(^{3}\) - (\(x - 3\)).[\(x + 3\)]\(^{2}\) - 3\(x\)(1 - \(x\))
C = \(x^{3} - 3 x^{2} + 3 x - 1\)- (\(x^{2} - 9 \left.\right) \left(\right. x + 3 \left.\right)\) - 3\(x\) + 3\(x^{2}\)
C = \(x^{3}\) - 3\(x^{2}\) + 3\(x - 1\) - \(x^{3}\) - 3\(x^{2}\) + 9\(x + 27\) - 3\(x\) + 3\(x^{2}\)
C = (\(x^{3}\) - \(x^{3}\)) - (3\(x^{2}\) + 3\(x^{2} - 3 x^{2}\)) + (3\(x\) - 3\(x + 9 x\)) + (-1 + 27)
C = 0 - (6\(x^{2}\) - 3\(x^{2}\)) + (0 + 9\(x\)) + 26
= - 3\(x^{2}\) + 9\(x\) + 26