Toán Lớp 10: (1-3x)^2-2x(x-3)=29
(x-2)^2-2×(x+2)×(1-2x)-42=3x
(2x+3)^2-(4+x)^2+8=0
3x^2-(3-x)×(4x+1)=47
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 10: (1-3x)^2-2x(x-3)=29
(x-2)^2-2×(x+2)×(1-2x)-42=3x
(2x+3)^2-(4+x)^2+8=0
3x^2-(3-x)×(4x+1)=47
Comments ( 2 )
a){\left( {1 – 3x} \right)^2} – 2x\left( {x – 3} \right) = 29\\
\Leftrightarrow 1 – 6x + 9{x^2} – 2{x^2} + 6x = 29\\
\Leftrightarrow 7{x^2} = 28\\
\Leftrightarrow {x^2} = 4\\
\Leftrightarrow x = 2;x = – 2\\
Vậy\,x = 2;x = – 2\\
b){\left( {x – 2} \right)^2} – 2\left( {x + 2} \right)\left( {1 – 2x} \right) – 42 = 3x\\
\Leftrightarrow {x^2} – 4x + 4 – 2\left( { – 2{x^2} – 3x + 2} \right) – 42 – 3x = 0\\
\Leftrightarrow {x^2} – 4x + 4 + 4{x^2} + 6x – 4 – 42 – 3x = 0\\
\Leftrightarrow 5{x^2} – x – 42 = 0\\
\Leftrightarrow 5{x^2} – 15x + 14x – 42 = 0\\
\Leftrightarrow \left( {x – 3} \right)\left( {5x + 14} \right) = 0\\
\Leftrightarrow x = 3;x = \dfrac{{ – 14}}{5}\\
Vậy\,x = 3;x = – \dfrac{{14}}{5}\\
c)\\
{\left( {2x + 3} \right)^2} – {\left( {4 + x} \right)^2} + 8 = 0\\
\Leftrightarrow 4{x^2} + 12x + 9 – {x^2} – 8x – 16 + 8 = 0\\
\Leftrightarrow 3{x^2} + 4x + 1 = 0\\
\Leftrightarrow 3{x^2} + 3x + x + 1 = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {3x + 1} \right) = 0\\
\Leftrightarrow x = – 1;x = – \dfrac{1}{3}\\
Vậy\,x = – 1;x = – \dfrac{1}{3}\\
d)3{x^2} – \left( {3 – x} \right).\left( {4x + 1} \right) = 47\\
\Leftrightarrow 3{x^2} – \left( { – 4{x^2} + 11x + 3} \right) = 47\\
\Leftrightarrow 3{x^2} + 4{x^2} – 11x – 3 – 47 = 0\\
\Leftrightarrow 7{x^2} – 11x – 50 = 0\\
\Leftrightarrow 7{x^2} + 14x – 25x – 50 = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {7x – 25} \right) = 0\\
\Leftrightarrow x = – 2;x = \dfrac{{25}}{7}\\
Vậy\,x = – 2;x = \dfrac{{25}}{7}
\end{array}$