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1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
a, \(\left(3x+2\right)^2-\left(2x-1\right)\left(2x+1\right)=5\left(x-2\right)^2\)
\(\Rightarrow9x^2+12x+4-\left(4x^2-1\right)=5\left(x^2-4x+4\right)\)
\(\Rightarrow9x^2+12x+4-4x^2-1=5x^2-20x+20\)
\(\Rightarrow9x^2-4x^2-5x^2+12x+20x=20+1-4\)
\(\Rightarrow32x=17\Rightarrow x=\dfrac{17}{32}\)
b, \(\left(x+2\right)^2-\left(x+3\right)\left(x-1\right)=5x\)
\(\Rightarrow x^2+4x+4-\left(x^2-x+3x-3\right)=5x\)
\(\Rightarrow x^2+4x+4-x^2+x-3x+3-5x=0\)
\(\Rightarrow-3x=-3-4\Rightarrow-3x=-7\Rightarrow x=\dfrac{7}{3}\)
c, \(\left(3x-1\right)\left(x-3\right)+\left(x-2\right)^2=\left(2x-5\right)^2\)
\(\Rightarrow3x^2-9x-x+3+x^2-4x+4=4x^2-20x+25\)
\(\Rightarrow3x^2+x^2-4x^2-9x-x-4x+20x=25-3-4\)
\(\Rightarrow6x=18\Rightarrow x=3\)
Chúc bạn học tốt!!!
(x-1)(x2+3x-2)-(x3-1)=0
<=>(x-1)(x2+3x-2)-(x-1)(x2+x+1)=0
<=>(x-1)(x2+3x-2-(x2+x+1))=0
<=>(x-1)(x2+3x-2-x2-x-1)=0
<=>(x-1)(2x-3)=0
<=>x-1=0 hay 2x-3=0
<=>x=1 hay x=\(\frac{3}{2}\)
- <=>(x-1)(x2+3x-2) - (x-1)(x2+x+1)=0
- <=>(x-1)(x2+3x-2-x2-x-1)=0
- <=>(x-1)(2x-3)=0
- <=>x-1=0 hoặc 2x-3=0
- <=>x=1 hoặc x=3/2
VẬY S=1;3/2 :)))))))))))))))))))))))))
a)\((x^2- 4).(x^2 - 10) = 72 Đặt x^2 - 7 = a(1), ta có (a+3)(a-3)=72 a^2-9=72 a^2=81 a=+-9 xét 2 trường hợp a = 9 và -9 khi thay vào (1) ta có..... tự lm nốt nha \)
b) nhóm x+1 vs x+4 và x+2 vs x+3 ta sẽ có (x2+5x+4)(x2+5x+6)(x+5)=40
\(\left(x^2-4\right)-\left(4x^2+4x+1\right)-2x+3x^2=0\)
\(\Leftrightarrow\left(x^2+3x^2-4x^2\right)+\left(-4x-2x\right)+\left(-4-1\right)=0\)
\(\Leftrightarrow-6x-5=0\Leftrightarrow x=-\frac{5}{6}\)
Vậy nghiệm phương trình là \(x=-\frac{5}{6}\)
\(\left(x-2\right)\left(x+2\right)-\left(2x+1\right)^2=x\left(2-3x\right)\)
\(\Leftrightarrow x^2-4-\left(4x^2+4x+1\right)=2x-3x^2\)
\(\Leftrightarrow x^2-4-4x^2-4x-1-2x+3x^2=0\)
\(\Leftrightarrow-5-6x=0\)
\(\Leftrightarrow-6x=5\Leftrightarrow x=\frac{-5}{6}\)