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\(1,\dfrac{x-2}{2}=3.\dfrac{1-3x}{6}\\ \Leftrightarrow\dfrac{x-2}{2}=\dfrac{1-3x}{2}\\ \Leftrightarrow x-2=1-3x\\ \Leftrightarrow4x=3\\ \Leftrightarrow x=\dfrac{3}{4}\)
2, mik có sửa đề vì đề của bn sai
ĐKXĐ:\(x\ne\pm\dfrac{1}{3}\)
\(\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}=\dfrac{5}{1-9x^2}\\ \Leftrightarrow\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{5}{\left(1-3x\right)\left(1+3x\right)}=0\\ \Leftrightarrow\dfrac{1-6x+9x^2-1-6x-9x^2-5}{\left(1+3x\right)\left(1-3x\right)}=0\\ \Rightarrow-12x-5=0\\ \Leftrightarrow x=-\dfrac{5}{12}\left(tm\right)\)
\(1,4x\left(1-x\right)-8=1-\left(4x^2+3\right)\\ \Leftrightarrow4x-4x^2-8=1-4x^2-3\\ \Leftrightarrow4x-4x^2-8-1+4x^2+3=0\\ \Leftrightarrow4x-6=0\\ \Leftrightarrow x=\dfrac{3}{2}\)
\(2,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\\ \Leftrightarrow\left(2-3x\right)\left(x+11\right)-\left(2-3x\right)\left(5x-2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(x+11-5x+2\right)=0\\ \Leftrightarrow\left(2-3x\right)\left(-4x+13\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
a, (x2 - 3)(2 + 4x)
= 2x2 + 4x3 - 6 - 12x
b, 6(3x + 5)(x - 4)
= (18x + 30)(x - 4)
= 18x2 - 72 + 30x - 120
= 18x2 + 30x - 192
\(Đk:\) \(x\ne1,x\ne2,x\ne3\)
\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)
\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Rightarrow0x-14=x-10\)
\(\Rightarrow x=-4\left(tmđk\right)\)
