tìm x
a, 6x:3-5=41
b, 1000000000:x=500
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a) 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/x.(x+1) = 499/500
1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/x - 1/x+1 = 499/500
1 - 1/x+1 = 499/500
1/x+1 = 1 - 499/500
1/x+1 = 1/500
x + 1 = 500
x = 500 - 1
x = 499
b) 1/1.3 + 1/3.5 + 1/5.7 + .... + 1/x.(x+2) = 20/41
1/2 . [ 2/1.3 + 2/3.5 + 2/5.7 + ... + 2/x.(x+2) ] = 20/41
1/2 . [ 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/x - 1/x+2 ] = 20/41
1/2 . [ 1 - 1/x+2 ) = 20/41
1 - 1/x+2 = 20/41 : 1/2
1 - 1/x+2 = 40/41
1/x+2 = 1 - 40/41
1/x+2 = 1/41
x + 2 = 41
x = 41 - 2
x = 39
a,575-(6x+70)=445
<=> 6x + 70 = 575 - 445
<=> 6x + 70 = 130
<=> 6x = 130 - 70
<=> 6x = 60
<=> x = 10
b, 720 : [ 41 - ( 2x - 5 ) ]= 23. 5
<=> 720 : [ 41 - ( 2x - 5 ) ] = 40
<=> 41 - ( 2x - 5 ) = 18
<=> 2x - 5 = 41 - 18
<=> 2x - 5 = 23
<=> 2x = 28
<=> x = 14
a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0
\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)
\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)
\(-5x-8=0\)
\(x=-\frac{8}{5}\)
Mai mik làm mấy bài kia sau
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
a) 6x : 3 - 5 = 41
2x - 5 = 41
2x = 41 + 5
2x = 46
x = 46 : 2
x = 23
b) 1 000 000 000 : x = 500
x = 1 000 000 000 : 500
x = 2 000 000
\(a)6x:3-5=41\)
\(\Rightarrow6x:3=46\)
\(\Rightarrow6x=138\)
\(\Rightarrow x=23\)
\(b)1000000000:x=500\)
\(\Rightarrow x=1000000000:500\)
\(\Rightarrow x=20000000\)