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\(P=\left(\dfrac{1}{ax-2}+\dfrac{1}{ax+2}+\dfrac{2ax}{a^2x^2+4}+\dfrac{4a^3x^3}{a^2x^4}\right)\cdot\dfrac{a^4x^4+16}{a^4x^4}\)
\(=\left(\dfrac{ax+2+ax-2}{a^2x^2-4}+\dfrac{2ax}{a^2x^2+4}+\dfrac{4a^3x^3}{a^4x^4}\right)\cdot\dfrac{a^4x^4+16}{a^4x^4}\)
\(=\left(\dfrac{2ax\left(a^2x^2+4\right)+2ax\left(a^2x^2-4\right)}{a^4x^4-16}+\dfrac{4a^3x^3}{a^4x^4}\right)\cdot\dfrac{a^4x^4+16}{a^4x^4}\)
\(=\left(\dfrac{4a^3x^3}{a^4x^4-16}+\dfrac{4a^3x^3}{a^4x^4}\right)\cdot\dfrac{a^4x^4+16}{a^4x^4}\)
\(=\dfrac{8a^7x^7-64a^3x^3}{a^4x^4\left(a^4x^4-16\right)}\cdot\dfrac{a^4x^4+16}{a^4x^4}=\dfrac{\left(8a^7x^7-64a^3x^3\right)\left(a^4x^4+16\right)}{a^8x^8\left(a^4x^4-16\right)}\)
\(=\dfrac{8a^3x^3\left(a^4x^4-8\right)\left(a^4x^4+16\right)}{a^8x^8\left(a^4x^4-16\right)}=\dfrac{8\left(a^4x^4-8\right)\left(a^4x^4+16\right)}{a^5x^5\left(a^4x^4-16\right)}\)
dài quá, làm từ từ nhé
1, \(\left(a-b\right)^2\left(2a-3b\right)-\left(b-a\right)^2\left(3a-5b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-b\right)^2\left(2a-3b-3a+5b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-b\right)^2\left(-a+2b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=-\left(a-b\right)^2\left(a-2b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-2b\right)\left[\left(a+b\right)^2-\left(a-b\right)^2\right]\)
\(=\left(a-2b\right)\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=4ab\left(a-2b\right)\)
2, \(x^4-4\left(x^2+5\right)-25=\left(x^2-25\right)-4\left(x^2+5\right)=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)
\(=\left(x^2-9\right)\left(x^2+5\right)=\left(x-3\right)\left(x+3\right)\left(x^2+5\right)\)
3,\(\left(2-x\right)^2+\left(x-2\right)\left(x+3\right)-\left(4x^2-1\right)=\left(x-2\right)^2+\left(x-2\right)\left(x+3\right)-\left(4x^2-1\right)\)
\(=\left(x-2\right)\left(x-2+x+3\right)-\left(2x-1\right)\left(2x+1\right)\)
\(=\left(x-2\right)\left(2x+1\right)-\left(2x-1\right)\left(2x+1\right)\)
\(=\left(x-2-2x+1\right)\left(2x+1\right)\)
\(=\left(-x-1\right)\left(2x+1\right)\)
4, câu này đề thiếu
5,\(16\left(xy+6\right)^2-\left(4x^2+y^2-25\right)^2=\left(4xy+24\right)^2-\left(4x^2+y^2-25\right)^2\)
\(=\left(4xy+24-4x^2-y^2+25\right)\left(4xy+24+4x^2+y^2-25\right)\)
\(=\left[49-\left(4x^2-4xy+y^2\right)\right]\left[\left(4x^2+4xy+y^2\right)-1\right]\)
\(=\left[49-\left(2x-y\right)^2\right]\left[\left(2x+y\right)^2-1\right]\)
\(=\left(7-2x+y\right)\left(7+2x-y\right)\left(2x+y-1\right)\left(2x+y+1\right)\)
a)\(\left(3x-1\right)^2-16=\left(3x-1-16\right)\left(3x-1+16\right)\)
\(=\left(3x-17\right)\left(3x+15\right)\)
c)\(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)
\(=\left(x-4\right)\left(x+14\right)\)
Aps dungj t/c a2 - b2 = ( a-b)(a+b)
A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)
B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Câu C) bạn xem lại đề nha mik tính ko đc
D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
\(\frac{a^2x^3-a^2}{ax^2+ax+a}=\frac{a^2\left(x^3-1\right)}{a\left(x^2+x+1\right)}=\frac{a^2\left(x-1\right)\left(x^2+x+1\right)}{a\left(x^2+x+1\right)}=a\left(x-1\right)=ax-a\)
\(\frac{12x^2-26x-16}{4x^2+4x+1}=\frac{\left(6x-16\right)\left(2x+1\right)}{\left(2x+1\right)^2}=\frac{6x-16}{2x+1}\)
\(\frac{\left(x+a\right)^2-x^2}{2x+a}=\frac{\left(x+a+x\right)\left(x+a-x\right)}{2x+a}=\frac{\left(2x+a\right)a}{2x+a}=a\)
Không ai trả lời luôn