b: \(x^4+4\)

\(=x^4+4x^2+4-4x^2\)

\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

Ta có: \(\frac{x^2}{x^2+2x+2}+\frac{x^2}{x^2-2x+2}=\frac{5\left(x^2-5\right)}{x^4+4}+\frac{25}{4}\)

=>\(\frac{x^2\left(x^2-2x+2\right)+x^2\left(x^2+2x+2\right)}{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}-\frac{5\left(x^2-5\right)}{\left(x^2+2x+2\right)\left(x^2-2x+2\right)}=\frac{25}{4}\)

=>\(\frac{x^4-2x^3+2x^2+x^4+2x^3+2x^2-5x^2+25}{x^4+4}=\frac{25}{4}\)

=>\(\frac{2x^4-x^2+25}{x^4+4}=\frac{25}{4}\)

=>\(25\left(x^4+4\right)=4\left(2x^4-x^2+25\right)\)

=>\(25x^4+100-8x^4+4x^2-100=0\)

=>\(17x^4+4x^2=0\)

=>\(x^2\left(17x^2+4\right)=0\)

=>\(x^2=0\)

=>x=0

a: \(x^4+4x^2+16\)

\(=x^4+8x^2+16-4x^2\)

\(=\left(x^2+4\right)_{}^2-\left(2x\right)^2=\left(x^2-2x+4\right)\cdot\left(x^2+2x+4\right)\)

\(\frac{x^2+2x}{\left(x+1\right)^2+3}-\frac{x^2-2x}{\left(x-1\right)^2+3}=\frac{16}{x^4+4x^2+16}\)

=>\(\frac{x^2+2x}{x^2+2x+4}-\frac{x^2-2x}{x^2-2x+4}=\frac{16}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}\)

=>\(\left(x^2+2x\right)\left(x^2-2x+4\right)-\left(x^2-2x\right)\left(x^2+2x+4_{}\right)=16\)

=>\(\left(x^2+2x\right)\left(x^2-2x\right)+4\left(x^2+2x\right)-\left(x^2-2x\right)\left(x^2+2x\right)-4\left(x^2-2x\right)=16\)

=>\(4\cdot\left(x^2+2x-x^2+2x\right)=16\)

=>4*4x=16

=>16x=16

=>x=1

\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)

bài 4 đâu bạn r =))

Thế mày làm đi

cho ít thôi thì làm

có ai giải ko

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

Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)

=>(x+5)(x-3)+8=x^2-1

=>x^2+2x-15+8=x^2-1

=>2x-7=-1

=>x=3(loại)

b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)

=>(x-4)(x+1)+x^2+3+5(x-1)=0

=>x^2-3x-4+x^2+3+5x-5=0

=>2x^2+2x-6=0

=>x^2+x-3=0

=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)

e: =>x^2-2x+1+2x+2=5x+5

=>x^2+3=5x+5

=>x^2-5x-2=0

=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)

g: (x-3)(x+4)*x=0

=>x=0 hoặc x-3=0 hoặc x+4=0

=>x=0;x=3;x=-4