Toán Lớp 8: h.(2x-1)^2=(x-3)^2
i.(2x-1)^2-(4x^2–1)=0
k.x^2(x^2+4)–x^2=4
m.x^4-x^3 +x^2 -x=0
n. x(2x-3)-3(3-2x)=0
p.4x^2–25–(2x-5)(2x+7) =0
q.x^3–8=(x-2)(x-12)
s.2(x+3) –x^2–3x=0
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Toán Lớp 8: h.(2x-1)^2=(x-3)^2
i.(2x-1)^2-(4x^2–1)=0
k.x^2(x^2+4)–x^2=4
m.x^4-x^3 +x^2 -x=0
n. x(2x-3)-3(3-2x)=0
p.4x^2–25–(2x-5)(2x+7) =0
q.x^3–8=(x-2)(x-12)
s.2(x+3) –x^2–3x=0
Comments ( 1 )
h){\left( {2x – 1} \right)^2} = {\left( {x – 3} \right)^2}\\
\Leftrightarrow {\left( {2x – 1} \right)^2} – {\left( {x – 3} \right)^2} = 0\\
\Leftrightarrow \left( {2x – 1 + x – 3} \right)\left( {2x – 1 – x + 3} \right) = 0\\
\Leftrightarrow \left( {3x – 4} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow x = \dfrac{4}{3};x = – 2\\
Vậy\,x = \dfrac{4}{3};x = – 2\\
i){\left( {2x – 1} \right)^2} – \left( {4{x^2} – 1} \right) = 0\\
\Leftrightarrow 4{x^2} – 4x + 1 – 4{x^2} + 1 = 0\\
\Leftrightarrow 4x = 2\\
\Leftrightarrow x = \dfrac{1}{2}\\
Vậy\,x = \dfrac{1}{2}\\
k){x^2}\left( {{x^2} + 4} \right) – {x^2} = 4\\
\Leftrightarrow {x^2}\left( {{x^2} + 4} \right) – {x^2} – 4 = 0\\
\Leftrightarrow \left( {{x^2} + 4} \right)\left( {{x^2} – 1} \right) = 0\\
\Leftrightarrow {x^2} – 1 = 0\\
\Leftrightarrow x = 1;x = – 1\\
Vậy\,x = 1;x = – 1\\
m){x^4} – {x^3} + {x^2} – x = 0\\
\Leftrightarrow {x^3}\left( {x – 1} \right) + x\left( {x – 1} \right) = 0\\
\Leftrightarrow x\left( {x – 1} \right).\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow x = 0;x = 1\\
Vậy\,x = 0;x = 1\\
n)x\left( {2x – 3} \right) – 3\left( {3 – 2x} \right) = 0\\
\Leftrightarrow \left( {2x – 3} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow x = \dfrac{3}{2};x = – 3\\
Vậy\,x = \dfrac{3}{2};x = – 3\\
p)4{x^2} – 25 – \left( {2x – 5} \right)\left( {2x + 7} \right) = 0\\
\Leftrightarrow \left( {2x – 5} \right)\left( {2x + 5} \right) – \left( {2x – 5} \right)\left( {2x + 7} \right) = 0\\
\Leftrightarrow \left( {2x – 5} \right)\left( {2x + 5 – 2x – 7} \right) = 0\\
\Leftrightarrow \left( {2x – 5} \right)\left( { – 2} \right) = 0\\
\Leftrightarrow x = \dfrac{5}{2}\\
Vậy\,x = \dfrac{5}{2}\\
q){x^3} – 8 = \left( {x – 2} \right)\left( {x – 12} \right)\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right) – \left( {x – 2} \right)\left( {x – 12} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + 2x + 4 – x + 12} \right) = 0\\
\Leftrightarrow \left( {x – 2} \right)\left( {{x^2} + x + 16} \right) = 0\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
s)2\left( {x + 3} \right) – {x^2} – 3x = 0\\
\Leftrightarrow 2\left( {x + 3} \right) – x\left( {x + 3} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2 – x} \right) = 0\\
\Leftrightarrow x = – 3;x = 2\\
Vậy\,x = – 3;x = 2
\end{array}$