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\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
<=> \(1-6x+9x^2-\left(9x^2-17x-2\right)=\left(9x^2-4\right)-\left[3\left(x+3\right)\right]^2\)
<=> \(1-6x+9x^2-9x^2+17x+2=9x^2-4-\left(3x+9\right)^2\)
<=> \(3+11x=\left(3x-3x-9\right)\left(3x+3x+9\right)-4\)
<=> \(3+4+11x=-9\left(6x+9\right)\)
<=> \(7+11x=-9.3\left(2x+3\right)\)
<=> \(7+11x=-27\left(2x+3\right)\)
<=> \(7+11x+27\left(2x+3\right)=0\)
<=> \(7+11x+54x+81=0\)
<=> \(65x=-88\)
<=> \(x=-\frac{88}{65}\)
(1 - 3x)2 - (x - 2)(9x + 1) = (3x - 4)(3x + 4) - 9(x + 3)2
⇌ 1 - 6x + 9x2 - 9x2 - x + 18x + 2 = 9x2 - 16 - 9x2 - 54x - 81 ⇌ 65x = -100 ⇌ x = \(-\frac{20}{13}\)\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
\(\Rightarrow1-6x+9x^2-9x^2+18x-x-2=9x^2-16-9x^2-6x-9\)
\(\Rightarrow\left(-6x+18x-x+6x\right)+\left(9x^2-9x^2-9x^2+9x^2\right)=-1+2-16-9\)
\(\Rightarrow17x=-24\)
\(\Rightarrow x=-\dfrac{24}{17}.\)
Vậy \(x=-\dfrac{24}{17}.\)
\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
\(\Rightarrow1-6x+9x^2-x\left(9x+1\right)+2\left(9x+1\right)=9x^2-16-9\left(x^2+6x+9\right)\)\(\Rightarrow1-6x+9x^2-9x^2-x-18x-2=9x^2-16-9x^2-54x-81\)\(\Rightarrow-1-24x=97-54x\)
\(\Rightarrow-1-24x-97+54x=0\)
\(\Rightarrow-98x+20x=0\)
\(\Rightarrow x=\dfrac{49}{10}\)
\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
\(\Leftrightarrow1-6x+9x^2-\left[x\left(9x+1\right)-2\left(9x+1\right)\right]=9x^2-16-9\left(x^2+6x+9\right)\)\(\Leftrightarrow1-6x+9x^2-\left(9x^2+x-18x-2\right)=9x^2-16-9x^2-54x-81\)\(\Leftrightarrow1-6x+9x^2-9x^2+x-18x-2=9x^2-16x-9x^2-54x-81\)\(\Leftrightarrow-1-24x=70x-81\)
\(\Leftrightarrow-1-24x-70x+81=0\)
\(\Leftrightarrow80-94x=0\)
\(\Leftrightarrow94x=80\Leftrightarrow x=\dfrac{40}{47}\)
a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
a: \(\Leftrightarrow\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2=4\)
\(\Leftrightarrow\left(3x+1-3x-5\right)^2=4\)
=>16=4(vô lý)
c: \(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6=7\)
=>18x+16=7
=>18x=-9
hay x=-1/2
Câu hỏi của Bảo Ngọc cute - Toán lớp 8 | Học trực tuyến :))
(1-3x2)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2
⇒1-3x2-(9x2+x-18x-2)=9x2-16-9(x2+6x+9)
⇒1-3x2-(9x2-17x-2)= -56x-97
⇒1-3x2-9x2+17x+2=-56x-97
⇒3-12x2+17x=-56x-97
⇒3-12x2+17x+56x+97=0
⇒-12x2+73x+100=0
⇒-(12x2-73x-100)=0