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\(\text{ĐKXĐ: }3x-2\ne0\text{ và }2+3x\ne0\)
\(\Leftrightarrow x\ne\frac{2}{3}\text{ và }x\ne-\frac{2}{3}\)
\(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)
\(\Leftrightarrow\frac{\left(3x+2\right)^2}{\left(3x-2\right)\left(3x+2\right)}-\frac{6.\left(3x-2\right)}{\left(3x-2\right)\left(3x+2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow9x^2+12x+4-18x+12=9x^2\)
\(\Leftrightarrow-6x+16=0\)
\(\Leftrightarrow x=\frac{8}{3}\)
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
a) (x2+x-6)(x2+9x+14) = 300
<=> (x-2)(x+3)(x+2)(x+7) - 300 = 0
<=> [(x-2)(x+7)][(x+2)(x+3)] - 300 = 0
<=> (x2-5x-14)(x2+5x+6) - 300 = 0
Đặt x2 + 5x - 14 = a
<=> a(a+20) - 300 = 0
<=> a2 + 20a - 300 = 0
<=> a2 + 20a + 100 - 400 = 0
<=> (a+10)2 - 202 = 0
<=> (a-10)(a+30) = 0
<=> \(\left[{}\begin{matrix}a=10\\a=-30\end{matrix}\right.\)
Với a = 10, ta có:
x2 + 5x - 14 = 10
=> x2 + 5x - 24 = 0
=> (x-3)(x+8) = 0
=> \(\left[{}\begin{matrix}x=3\\x=-8\end{matrix}\right.\)
Với a = -30, ta có:
x2 + 5x - 14 = -30
=> x2 + 5x + 16 = 0 (vn)
Vậy nghiệm pt x = 3; x = -8
b) (2x-5)(3x+1) = 4x2 - 25
<=> (2x-5)(3x+1) = (2x-5)(2x+5)
<=> (2x-5)(3x+1-2x-5) = 0
<=> (2x-5)(x-4) = 0
<=> \(\left[{}\begin{matrix}2x-5=0\\x-4=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=4\end{matrix}\right.\)
Vậy...
Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(x^2-3x+9-\frac{3}{x}+\frac{1}{x^2}=0\)
\(\Leftrightarrow x^2+\frac{1}{x^2}-3\left(x+\frac{1}{x}\right)+9=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
pt trở thành: \(t^2-2-3t+9=0\)
\(\Leftrightarrow t^2-3t+7=0\) (vô nghiệm)
Vậy pt đã cho vô nghiệm
mk giải từng nha == tại vì mk sợ nhiều qus bị troll
\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(27x^3+18x^2+12x-18x^2-12x-8-3x\left(9x^2-3x+1\right)+\left(9x^2-3x+1\right)=x-4\)
\(27x^3-8-3\left(9x^2-3x+1\right)+9x^2-3x+1=x-4\)
\(27x^3-7-3x\left(9x^2-3x+1\right)+9x^2-3x=x-4\)
\(27x^3-7-27x^3+9x^2-3x+9x^2-3x=x-4\)
\(-7+18x^2-6x=x-4\)
\(3-18x^2+7x=0\)
\(x=\frac{-7+\sqrt{265}}{-36};\frac{-7-\sqrt{265}}{-36}\)
\(9\left(2x+1\right)=4\left(x-5\right)^2\)
\(18x+9=4x^2-40x+100\)
\(18x+9-4x^2+40x-100=0\)
\(58x-91-4x^2=0\)
\(x=\frac{29-3\sqrt{53}}{4};\frac{29+3\sqrt{53}}{4}\)
Câu hỏi của Trịnh Minh Châu - Toán lớp 8 - Học toán với OnlineMath
\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{9x}{x^2-7x+10}=10\)
\(\Leftrightarrow\frac{3x^2-15x-x^2+2x+9x}{\left(x-2\right)\left(x-5\right)}=10\)
\(\Leftrightarrow2x^2-4x=10x^2-70x+100\)
\(\Leftrightarrow8x^2-66+100=0\)
\(\Leftrightarrow4x^2-33x+50=0\)
\(\Leftrightarrow4x\left(x-2\right)-25\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x-25\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{25}{4}\end{matrix}\right.\)
b) [(x-7)(x-2)][(x-4)(x-5)]=72
<=> (x2-9x+14)(x2-9x+20)=72
Đặt x2-9x+17=a
=> (a+3)(a-3)=72
<=> a2-9=72
<=> a2=81
=> a=\(\left\{9;-9\right\}\)
TH1: a=9
=> x2-9x+17=9
<=> x2-9x+8=0
<=> (x-1)(x-8)=0
=> x=\(\left\{1;8\right\}\)
TH2: a=-9
=> x2-9x+17=-9
<=> x2-9x+26=0
<=> x2-9x+20,25+5,75=0
<=> (x-4,5)2+5,75=0
=> x\(\in\varnothing\)
Vậy x=\(\left\{1;8\right\}\)
\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)=\left(2-3x\right)\left(2+3x\right)\)
\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)-\left(2-3x\right)\left(2+3x\right)=0\)
\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)
\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2+3x=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-\dfrac{2}{3};-\dfrac{3}{2}\right\}\)
Phương trình trên tương đương:
(5-x)(2+3x)=(2-3x)(2+3x)
(5-x)(2+3x)-(2-3x)(2+3x)=0
Đặt 2+3x làm nhân tử chung rồi giải pt tích rồi kết luận