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

bài 4 đâu bạn r =))

a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

\(\Leftrightarrow9x+7=16\)

\(\Leftrightarrow9x=9\)

hay x=1

a,2/3 + 4/-5 x 20/16 =-1/3

b,(1/3+4/6).(2/7+9/14)=13/14

c,(2/3 - 3/4).(1/2- -3/5)=-11/120

a,2/3 + 4/-5 x 20/16
=2/3-1
=-1/3
b,(1/3+4/6).(2/7+9/14)
=1.13/14
=13/14
c,(2/3 - 3/4).(1/2- -3/5)
=-1/12.-1/10
=1/120

\(a,\Rightarrow\left(4x-1\right)^2=25=5^2=\left(-5\right)^2\\ \Rightarrow\left[{}\begin{matrix}4x-1=5\\4x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\\ b,\Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:9=16=2^4\Rightarrow x=4\\ c,\Rightarrow3^{2x+3}=3^{2\left(x+3\right)}\\ \Rightarrow2x+3=2x+6\Rightarrow0x=3\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)

Th1: 2x+3 ≥ 0
Khi đó: |2x+3| =x+2
 (2x+3)= x+2
- 2x+3= x+2
-2x-x= 2-3
 x= -1
Th2: 2x+3 < 0
Khi đó: |2x+3|=x+2
 -(2x+3) = x +2
 -2x-3 = x+2
 -3x = 5
 x=-5/3

Vậy x= -1

      x= -5/3

Lớp 6 cugx học dạng v nè

Câu b nha 

a: Khi x=16 thì \(A=\dfrac{2\cdot\sqrt{16}}{\sqrt{16}+3}=\dfrac{2\cdot4}{4+3}=\dfrac{8}{7}\)

b: P=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{7\sqrt{x}+3}{9-x}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x+5\sqrt{x}+6}{x-9}\)

a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)

\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)

\(\Leftrightarrow-10x^2-10x=0\)

\(\Leftrightarrow-10x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c: Ta có: \(x^3+3x^2+3x+28=0\)

\(\Leftrightarrow\left(x+1\right)^3=-27\)

\(\Leftrightarrow x+1=-3\)

hay x=-4

a: \(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)

c: \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)

\(\dfrac{1}{8}x^3-64=\left(\dfrac{1}{2}x-4\right)\left(\dfrac{1}{4}x^2+2x+16\right)\)

d: \(=\left(2x+5y\right)^3\)

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