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5 tháng 11 2021

giúp mik với

 

\(\Leftrightarrow2x+11=17\)

hay x=3

28 tháng 10 2021

-2.(2x+11) =80 -144

      -4x -22 = -64

          -4x    = -64 +22

            -4x   = -42

               x    =10,5

28 tháng 10 2021

\(114-4x-22-80=0\)

\(12-4x=0\)

\(x=3\)

28 tháng 10 2021

\(114-2\cdot\left(2x+11\right)=80\)

\(\Rightarrow2x+11=17\Rightarrow x=3\)

29 tháng 9 2023

\(a,210-5\left(x-11\right)=200\\ \Rightarrow5\left(x-11\right)=10\\ \Rightarrow x-11=2\\ \Rightarrow x=13.\\ b,\left(2x+1\right):7=2^2+3^2\\ \Rightarrow\left(2x+1\right):7=4+9\\ \Rightarrow\left(2x+1\right):7=13\\ \Rightarrow2x+1=91\\ \Rightarrow2x=90\\ \Rightarrow x=45.\\ c,450:\left[41-\left(2x-5\right)\right]=3^2.5\\ \Rightarrow450:\left[41-\left(2x-5\right)\right]=45\\ \Rightarrow41-\left(2x-5\right)=10\\ \Rightarrow2x-5=31\\ \Rightarrow2x=36\\ \Rightarrow x=18.\\ d,\left(5x-9\right).7+3=80\\ \Rightarrow\left(5x-9\right).7=77\\ \Rightarrow5x-9=11\\ \Rightarrow5x=20\\ \Rightarrow x=4.\)

29 tháng 9 2023

a) \(210-5\left(x-11\right)=200\)

\(5.\left(x-11\right)=210-200\)

\(5.\left(x-11\right)=10\)

\(x-11=10:5\)

\(x-11=2\)

\(x=2+11\)

\(x=13\)

b) \(\left(2x+1\right):7=2^2+3^2\)

\(\left(2x+1\right):7=13\)

\(2x+1=13.7\)

\(2x=91-1\)

\(2x=90\)

\(x=90:2\)

\(x=45\)

c) \(450:\left[41-\left(2x-5\right)\right]=3^2.5\)

\(450:\left[41-\left(2x-5\right)\right]=45\)

\(\left[41-\left(2x-5\right)\right]=450:45\)

\(\left[41-\left(2x-5\right)\right]=10\)

\(2x-5=41-10\)

\(2x-5=31\)

\(2x=31+5\)

\(2x=36\)

\(x=36:2\)

\(x=18\)

d) \(\left(5x-9\right).7+3=80\)

\(\left(5x-9\right).7=80-3\)

\(5x-9=77:7\)

\(5x-9=11\)

\(5x=11+9\)

\(5x=20\)

\(x=20:5\)

\(x=4\)

11) Ta có: \(a^6+a^4+a^2b^2+b^4-b^6\)

\(=a^6-b^6+a^4+a^2b^2+b^4\)

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

\(=\left(a^4+a^2b^2+b^4\right)\left(a^2-b^2+1\right)\)

12) Ta có: \(x^3+3xy+y^3-1\)

\(=\left(x^3+3x^2y+3xy^2+y^3-1\right)-3x^2y-3xy^2+3xy\)

\(=\left[\left(x+y\right)^3-1\right]-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left[x^2+2xy+y^2+x+y+1\right]-3xy\left(x+y-1\right)\)

\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)

14) Ta có: \(x^8+x+1\)

\(=x^8+x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3+x^2-x^2+x+1\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

15) Ta có: \(x^8+3x^4+4\)

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

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

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

7 tháng 2 2022

a) \(7x=63\Leftrightarrow x=9\)

b) \(x=682\)

c) \(x=80\)

d) \(1043\)

7 tháng 2 2022

a x * 7=63

x=63:7

x=9

b 8 * x=640

x=640:8

x=80

c x=341 * 2

x=682

d x:7=149

x=149 * 7

x =1043

a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)

=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)

=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22

=>-x^2+59x+14-8x^2+5x+22=0

=>-9x^2+54x+36=0

=>x^2-6x-4=0

=>\(x=3\pm\sqrt{13}\)

b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)

=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)

=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32

=>x^2+6x+19=x^2+4x-32

=>2x=-51

=>x=-51/2