Tìm x:
2 (x-5) -3 (x+7)=14
Ai nhanh, mk tick
rõ ràng nha
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45 - ( x + 45 - 3 ) = l -7 + 3 l
45 - x - 45 + 3 = 4
45 - x - 45 = 4 - 3
45 - x - 45 = 1
45 - x = 1 + 45
45 - x = 46
=> x = 45 - 46
=> x = -1
Vậy x = -1
(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)(8x−3)(3x+2)−(4x+7)(x+4)=(2x+1)(5x−1)
20x2−16x−34=10x2+3x−120x2−16x−34=10x2+3x−1
10x2−19x−33=010x2−19x−33=0
(10x+11)(x−3)=0
chỉ bt lm con b thoy
..army,,,,,,,,,,
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
\(\Leftrightarrow3x^2-12x=3x^2-17x+20+2\)
\(\Leftrightarrow3x^2-12x=3x^2-17x+22\left(3x^2-17x\right)\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\frac{22}{5}\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x+1\)
\(\Leftrightarrow20x^2-16x-33=10x^2+3x\)
\(\Leftrightarrow20x^2-16x-33=10x^2+3x-3x\)
\(\Leftrightarrow20x^2-16x-33=10x^2\)
\(\Leftrightarrow20x^2-16x-33=10x^2-10x^2\)
\(\Leftrightarrow20x^2-16x-33=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-\frac{11}{10}\end{cases}}\)
a, (x + 2) + (x + 4) + (x + 6) + ... + (x + 50) = 750
=> x + 2 + x + 4 + x + 6 + ... + x + 50 = 750
=> (x + x + x + ... + x) + (2 + 4 + 6 + ... + 50) = 750
=> 25x + (50 + 2).25 : 2 = 750
=> 25x + 52.25 : 2 = 750
=> 25x + 650 = 750
=> 25x = 100
=> x = 4
a) ( x+x+...+x)+(2+4+6+...+50)= 750
( x*25)+ (50+2)*25:2 = 750
(x*25)+ 650 = 750
x* 25 = 750 - 650 = 100
x = 100 :25 = 4
Bài 1:
1 - 2 + 3 - 4 + .... + 2009 - 2010
= (1 - 2) + (3 - 4) + ... + (2009 - 2010)
= -1 . 1005
= -1005
Bài 2:
a) (15 - x) - (-x + 12) = 7 - (-5 + x)
=> 15 - x + x - 12 = 7 + 5 - x
=> -x + x + x = 7 + 5 - 15 + 12
=> x = 9
b) (x - 5)4 = (x - 5)6
=> x - 5 = 1 hoặc x - 5 = 0
=> x = 6 hoặc x = 5
c) (x + 1) + (x + 3) + ... + (x + 99) = 0
=> (x . 50) + (1 + 3 + ... + 99) = 0
=> (x . 50) + 2500 = 0
=> x . 50 = -2500
=> x = -50
\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
<=>\(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}.\left(1+5+5^3\right)}{131}\)
<=>\(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)
<=>\(7^x=5^{2x}\)<=>\(7^x=10^x\)<=>x=0
Vậy x=0
2(x-5)-3(x+7)=14
=> 2x-10-3x-21=14
=> x(2-3)-10-21=14
=> x.(-1)-31=14
=> -x=14+31
=> -x=45
=> x=-45
2 (x-5) -3 (x+7)=14
<=>2x-10-3x-21=14
<=>-x-31=14
<=>-x=45
<=>x=-45