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a: \(VT=x^2-25-x^2+4x-4-7x^2+x^3+1\)
\(=x^3-7x^2+4x-28\)
\(VP=x^3+9x^2+27x+27-x^3-9x^2=27x+27\)
=>\(x^3-7x^2-23x-55=0\)
=>\(x\in\left\{9.89\right\}\)
b: \(\Leftrightarrow4x^2+12x+9+x^2-1-5x^2-20x-20=-\left(x^2-4x-5\right)+x^2+8x+16\)
=>\(-8x-12=-x^2+4x+5+x^2+8x+16\)
=>-8x-12=12x+21
=>-20x=33
=>x=-33/20
\(a.\left(2x+5\right)\frac{6}{2}=75\\ \Leftrightarrow\left(2x+5\right)3=75\\ \Leftrightarrow6x+15=75\\\Leftrightarrow 6x=75-15\\\Leftrightarrow 6x=60\\ \Leftrightarrow x=10\)
\(b.\frac{x-3}{5}=6-\frac{1-2x}{3}\\ \Leftrightarrow\frac{3\left(x-3\right)}{15}=\frac{6.15}{15}-\frac{5\left(1-2x\right)}{15}\\ \Leftrightarrow3\left(x-3\right)=6.15-5\left(1-2x\right)\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-9-90+5-10x=0\\ \Leftrightarrow-7x-94=0\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\frac{-94}{7}\)
\(c.\frac{2x}{3}+\frac{2x-1}{6}=\frac{4-x}{3}\\ \Leftrightarrow\frac{2x.2}{6}+\frac{2x-1}{6}=\frac{2\left(4-x\right)}{6}\\ \Leftrightarrow2x.2+2x-1=2\left(4-x\right)\\ \Leftrightarrow4x+2x-1=8-2x\\ \Leftrightarrow4x+2x-1-8+2x=0\\ \Leftrightarrow8x-9=0\\ \Leftrightarrow8x=9\\ \Leftrightarrow x=\frac{9}{8}\)
\(d.\frac{x-1}{2}+\frac{x-1}{4}=\frac{1-x}{3}\\ \Leftrightarrow\frac{6\left(x-1\right)}{12}+\frac{3\left(x-1\right)}{12}=\frac{4\left(1-x\right)}{12}\\ \Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=4\left(1-x\right)\\ \Leftrightarrow6x-6+3x-3=4-4x\\ \Leftrightarrow6x-6+3x-3-4+4x=0\\ \Leftrightarrow13x-13=0\\ \Leftrightarrow13x=13\\ \Leftrightarrow x=1\)
\(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\left(x\ne\pm2\right)\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{2x^2+4}{x^2-4}=\frac{2x^2+4}{x^2-4}\)
Vậy phương trình này có vô số nghiệm x thỏa mãn trừ x khác 2 và -2
\(a)\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\\ \Leftrightarrow15x-63x^2-15+63x+63x^2-35x+36x-20=44\\ \Leftrightarrow79x-35=44\\ \Leftrightarrow79x=79\Rightarrow x=1\)
\(b)\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\\ \Leftrightarrow x^2+2x+x+2\left(x+5\right)-x^3-8x^2=27\\ \Leftrightarrow x^2\left(x+5\right)+2x\left(x+5\right)+x\left(x+5\right)+2\left(x+5\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+5x^2+2x^2+10x+x^2+5x+2x+10-x^3-8x^2=27\\ \Leftrightarrow17x+10=27\\ \Leftrightarrow17x=17\Rightarrow x=1\)