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3735
2436
`13) x(x-3) -x+3=0`
`<=> x(x-3) -(x-3) =0`
`<=> (x-3)(x-1) =0`
`<=> [(x-3=0),(x-1=0):}`
`<=> [(x=3),(x=1):}`
Vậy `x in {3;1}`
`14) 2x(x-1) -x+1=0`
`<=> 2x(x-1) -(x-1) =0`
`<=> (x-1)(2x-1) =0`
`<=> [(x-1=0),(2x-1=0):}`
`<=> [(x=1),(x=1/2):}`
Vậy `x in {1;1/2}`
`15) x(x+3) -2x-6=0`
`<=> x(x+3)-(2x+6) =0`
`<=> x(x+3) -2(x+3) =0`
`<=> (x+3)(x-2) =0`
`<=> [(x+3=0),(x-2=0):}`
`<=> [(x=-3),(x=2):}`
Vậy `x in {-3;2}`
`16) 2x(x+5)-x-5=0`
`<=> 2x(x+5) -(x+5) =0`
`<=> (x+5)(2x-1) =0`
`<=> [(x+5=0),(2x-1=0):}`
`<=> [(x=-5),(x=1/2):}`
Vậy `x in {-5;1/2}`
`17) x(x-3) +2x-6=0`
`<=> x(x-3) +(2x-6) =0`
`<=> x(x-3) +2(x-3) =0`
`<=> (x-3)(x+2) =0`
`<=> [(x-3=0),(x+2=0):}`
`<=> [(x=3),(x=-2):}`
Vậy `x in {3;-2}`
`18) x(x-1) +2x-2=0`
`<=> x(x-1) +(2x-2) =0`
`<=> x(x-1) +2(x-1) =0`
`<=> (x-1)(x+2) =0`
`<=> [(x-1=0),(x+2=0):}`
`<=> [(x=1),(x=-2):}`
Vậy `x in {1;-2}`
`\ttcolor{#00FF00}{#}\ttcolor{#00EE00}{b}\ttcolor{#00DD00}{u}\ttcolor{#00CC00}\ttcolor{#00BB00}{i}\ttcolor{#00AA00}{g}\ttcolor{#009900}{i}\ttcolor{#008800}{a}\ttcolor{#007700}{p}\ttcolor{#006600}{h}\ttcolor{#005500}{o}\ttcolor{#004400}{n}\ttcolor{#003300}\ttcolor{#002200}{g}\ttcolor{#001100}{999}`
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1343
803
`13)` `x(x - 3) - x + 3 = 0`
`x(x - 3) - (x - 3) = 0`
`(x - 3)(x - 1) = 0`
`@ x - 3 = 0`
`-> x = 3`
`@ x - 1 = 0`
`-> x = 1`
Vậy `x \in {3; 1}`
`14)` `2x(x - 1) - x + 1 = 0`
`2x(x - 1) - (x - 1) = 0`
`(x - 1)(2x - 1) = 0`
`@ x - 1 = 0`
`-> x = 1`
`@ 2x - 1 = 0`
`-> 2x = 1`
`-> x = 1/2`
Vậy `x \in {1; 1/2}`
`15)` `x(x + 3) - 2x - 6 = 0`
`x(x + 3) - 2(x + 3) = 0`
`(x + 3)(x - 2) = 0`
`@ x + 3 = 0`
`-> x = -3`
`@ x - 2 = 0`
`-> x = 2`
Vậy `x \in {-3; 2}`
`16)` `2x(x + 5) - x - 5 = 0`
`2x(x + 5) - (x + 5) = 0`
`(x + 5)(2x - 1) = 0`
`@ x + 5 = 0`
`-> x = -5`
`@ 2x - 1 = 0`
`-> 2x = 1`
`-> x = 1/2`
Vậy `x \in {-5; 1/2}`
`17)` `x(x - 3) + 2x - 6 = 0`
`x(x - 3) + 2(x - 3) = 0`
`(x - 3)(x + 2) = 0`
`@ x - 3 = 0`
`-> x = 3`
`@ x + 2 = 0`
`-> x = -2`
Vậy `x \in {3; -2}`
`18)` `x(x - 1) + 2x - 2 = 0`
`x(x - 1) + 2(x - 1) = 0`
`(x - 1)(x + 2) = 0`
`@ x - 1 = 0`
`-> x = 1`
`@ x + 2 = 0`
`-> x = -2`
Vậy `x \in {1; -2}`
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