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a,\(4x\left(2x+3\right)-x\left(8x-1\right)=5\left(x+2\right)\)
\(< =>8x^2+12x-8x^2+x=5x+10\)
\(< =>13x=5x+10< =>8x=10\)
\(< =>x=\frac{10}{8}=\frac{5}{4}\)
b, \(\left(3x-5\right)\left(3x+5\right)-x\left(9x-1\right)=4\)
\(< =>9x^2-25-9x^2+x=4\)
\(< =>x=4+29=33\)
c,\(3-4x\left(25-2x\right)=8x^2+x-300\)
\(< =>3-100x+8x^2=8x^2+x-300\)
\(< =>x+100x=3+300\)
\(< =>101x=303< =>x=\frac{303}{101}=3\)
d,\(2\left(1-\frac{3x}{5}\right)-\frac{2+3x}{10}=7-\frac{3\left(2x+1\right)}{4}\)
\(< =>2-\frac{6x}{5}-\frac{2+3x}{10}=7-\frac{6x+3}{4}\)
\(< =>-\frac{24x}{20}-\frac{4+6x}{20}+\frac{30x+15}{20}=5\)
\(< =>\frac{30x-6x-24x+15-4}{20}=5\)
\(< =>\frac{11}{5}=5< =>11=25\)(vo li)
c: Ta có: \(\left(2x-3\right)^2-\left(2x-3\right)\left(x-10\right)=7\)
\(\Leftrightarrow4x^2-12x+9-2x^2+20x+3x-30=7\)
\(\Leftrightarrow11x=28\)
hay \(x=\dfrac{28}{11}\)
d: Ta có: \(\left(3x-4\right)^2-9\left(x-3\right)\left(x+3\right)=8\)
\(\Leftrightarrow9x^2-24x+16-9x^2+81=8\)
\(\Leftrightarrow-24x=-89\)
hay \(x=\dfrac{89}{24}\)
f: Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\)
hay \(x=-\dfrac{1}{8}\)
a: =>x^2-25-x^2-3x=10
=>-3x=35
=>x=-35/3
b: =>4x^2-9-4(x^2+4x+4)=5
=>4x^2-9-4x^2-16x-16-5=0
=>-16x-30=0
=>x=-15/8
c: =>9x^2+45x-9x^2+4=7
=>45x=3
=>x=1/15
d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8
=>8x=7
=>x=7/8
\(2x+\frac{2}{5}+\frac{3}{10}< 3x-\frac{2}{4}\)
\(=>2x+0,4+0,3< 3x-0,5\)
\(=>2x+0,7< 3x-0,5\)
\(=>-x=-1,2\)
\(=>x=1,2\)
e lam lai cho minh voi