Đáp án `+` Giải thích các bước giải `!`
`to` Tìm `x:`
`a)`
`3x(2x+7)-(x+1)(6x-5) = 10`
`<=> 6x^2+21x-6x^2+5x-6x+5 = 10`
`<=> (6x^2-6x^2)+(21x+5x-6x)+5 = 10`
`<=> 20x = 5`
`<=> x = 1/4`
Vậy `S= {1/4}`
`b)`
`(x-4)^2-36 = 0`
`<=> (x-4)^2-6^2 = 0`
`<=> (x-4-6)(x-4+6) = 0`
`<=> (x-10)(x+2) = 0`
`<=>` \(\left[ \begin{array}{l}x-10=0\\x+2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=10\\x=-2\end{array} \right.\)
Vậy `S= {10; -2}`
Áp dụng: `a^2-b^2 = (a-b)(a+b)`
`c)`
`3(2x-1)-5(x-3) = 4`
`<=> 6x-3-5x+15 = 4`
`<=> (6x-5x)+(-3+15) = 4`
`<=> x+12 = 4`
`<=> x = -8`
Vậy `S= {-8}`
`d)`
`(2x-1)^2-(x+3)^2 = 0`
`<=> (2x-1-x-3)(2x-1+x+3) = 0`
`<=> (x-4)(3x+2) = 0`
`<=>` \(\left[ \begin{array}{l}x-4=0\\3x+2=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=4\\3x=-2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=4\\x=-\dfrac{2}{3}\end{array} \right.\)
Vậy `S= {4; -(2)/3}`
Áp dụng: `a^2-b^2 = (a-b)(a+b)`
`e)`
`(x-3)(x-7)-x^2 = 1`
`<=> x^2-7x-3x+21-x^2 = 1`
`<=> (x^2-x^2)+(-7x-3x)+21 = 1`
`<=> -10x = -20`
`<=> x = 2`
Vậy `S= {2}`
`f)`
`(x+2)^2-(x-2)(x-1) = 0`
`<=> x^2+4x+4-x^2+x+2x-2 = 0`
`<=> (x^2-x^2)+(4x+x+2x)+(4-2) = 0`
`<=> 7x+2 = 0`
`<=> 7x = -2`
`<=> x = -(2)/7`
Vậy `S= {-(2)/7}`
`g)`
`(x-1)^2-(x-2)(x+2) = 0`
`<=> x^2-2x+1-x^2+4 = 0`
`<=> (x^2-x^2)-2x+(1+4) = 0`
`<=> -2x+5 = 0`
`<=> -2x = -5`
`<=> x = 5/2`
Vậy `S= {5/2}`
Áp dụng:
`+)` `(a-b)^2 = a^2-2ab+b^2`
`+)` `a^2-b^2 = (a-b)(a+b)`
`h)`
`(10x+9). x-(5x-1)(2x+3) = 8`
`<=> 10x^2+9x-10x^2-15x+2x+3 = 8`
`<=> (10x^2-10x^2)+(9x-15x+2x)+3 = 8`
`<=> -4x = 5`
`<=> x = -(5)/4`
Vậy `S= {-(5)/4}`
