Toán Lớp 8: Tìm các biểu thức A,B biết
a) x²-x+A=B²
b) 4x²-A+1=B²
c) (5-3y)(5+3y)=A²-B²
d) (A-6xy+9y²)=B²
e) 2x²+8x+10=A²+B²
f) (3x+5)(x-1)=A²-B²
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 8: Tìm các biểu thức A,B biết
a) x²-x+A=B²
b) 4x²-A+1=B²
c) (5-3y)(5+3y)=A²-B²
d) (A-6xy+9y²)=B²
e) 2x²+8x+10=A²+B²
f) (3x+5)(x-1)=A²-B²
Comments ( 1 )
a){x^2} – x + A = {B^2}\\
\Leftrightarrow {x^2} – 2.x.\dfrac{1}{2} + \dfrac{1}{4} – \dfrac{1}{4} + A = {B^2}\\
\Leftrightarrow {\left( {x – \dfrac{1}{2}} \right)^2} + A – \dfrac{1}{4} = {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = \dfrac{1}{4}\\
B = x – \dfrac{1}{2}
\end{array} \right.\\
b)4{x^2} – A + 1 = {B^2}\\
\Leftrightarrow 4{x^2} – 4x + 1 + 4x – A = {B^2}\\
\Leftrightarrow {\left( {2x – 1} \right)^2} + 4x – A = {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = 4x\\
B = 2x – 1
\end{array} \right.\\
c)\left( {5 – 3y} \right)\left( {5 + 3y} \right) = {A^2} – {B^2}\\
\Leftrightarrow {5^2} – {\left( {3y} \right)^2} = {A^2} – {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = 5\\
B = 3y
\end{array} \right.\\
d)\left( {A – 6xy + 9{y^2}} \right) = {B^2}\\
\Leftrightarrow {x^2} – 6xy + 9{y^2} – {x^2} + A = {B^2}\\
\Leftrightarrow {\left( {x – 3y} \right)^2} + A – {x^2} = {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = {x^2}\\
B = x – 3y
\end{array} \right.\\
e)2{x^2} + 8x + 10 = {A^2} + {B^2}\\
\Leftrightarrow 2.\left( {{x^2} + 4x + 4} \right) + 2 = {A^2} + {B^2}\\
\Leftrightarrow 2.{\left( {x + 2} \right)^2} + 2 = {A^2} + {B^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = \sqrt 2 \left( {x + 2} \right)\\
B = \sqrt 2
\end{array} \right.\\
f)\left( {3x + 5} \right)\left( {x – 1} \right) = {A^2} – {B^2}\\
\Leftrightarrow 3{x^2} – 3x + 5x – 5 = {A^2} – {B^2}\\
\Leftrightarrow 3{x^2} + 2x – 5 = {A^2} – {B^2}\\
\Leftrightarrow {\left( {\sqrt 3 x} \right)^2} + 2.\sqrt 3 x.\dfrac{1}{{\sqrt 3 }} + \dfrac{1}{3} – \dfrac{1}{3} – 5 = {A^2} – {B^2}\\
\Leftrightarrow {\left( {\sqrt 3 x + \dfrac{1}{{\sqrt 3 }}} \right)^2} – \dfrac{{16}}{3} = {A^2} – {B^2}\\
\Leftrightarrow {\left( {\sqrt 3 x + \dfrac{1}{{\sqrt 3 }}} \right)^2} – {\left( {\dfrac{{4\sqrt 3 }}{3}} \right)^2}\\
\Leftrightarrow \left\{ \begin{array}{l}
A = \sqrt 3 x + \dfrac{1}{{\sqrt 3 }}\\
B = \dfrac{{4\sqrt 3 }}{3}
\end{array} \right.
\end{array}$