# 1 tìm x a(x-2)*(x+2)+x=2x+x^2+4 b x^2*(x-3)-x^3=x^2+4 c (x-2)*(x+3)=0 D,(x^n-2x^4)*x^2-x^n-1*(2x^3-3x-1)=0 2 , a=2x^4-61x^3-95x^2-31x+64 vs x=32 b x^5

1 tìm x
a(x-2)*(x+2)+x=2x+x^2+4
b x^2*(x-3)-x^3=x^2+4
c (x-2)*(x+3)=0
D,(x^n-2x^4)*x^2-x^n-1*(2x^3-3x-1)=0
2 , a=2x^4-61x^3-95x^2-31x+64 vs x=32
b x^5-2013 x^4+2015x^3-4023x^2 vs x=2012

### 0 thoughts on “1 tìm x a(x-2)*(x+2)+x=2x+x^2+4 b x^2*(x-3)-x^3=x^2+4 c (x-2)*(x+3)=0 D,(x^n-2x^4)*x^2-x^n-1*(2x^3-3x-1)=0 2 , a=2x^4-61x^3-95x^2-31x+64 vs x=32 b x^5”

1. Giải thích các bước giải:

B1:

$\begin{array}{l} a)(x – 2)(x + 2) + x = 2x + {x^2} + 4\\ \Leftrightarrow {x^2} – 4 + x = {x^2} + 2x + 4\\ \Leftrightarrow x = – 8\\ b){x^2}(x – 3) – {x^3} = {x^2} + 4\\ \Leftrightarrow {x^3} – 3{x^2} – {x^3} = {x^2} + 4\\ \Leftrightarrow 4{x^2} + 4 = 0\left( {vn,4{x^2} + 4 \ge 4 > 0,\forall x} \right)\\ c)\left( {x – 2} \right)\left( {x + 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x – 2 = 0\\ x + 3 = 0 \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = 2\\ x = – 3 \end{array} \right.\\ d)\left( {{x^n} – 2{x^4}} \right){x^2} – {x^{n – 1}}\left( {2{x^3} – 3x – 1} \right) = 0\\ \Leftrightarrow {x^{n + 2}} – 2{x^6} – 2{x^{n + 2}} + 3{x^n} + {x^{n – 1}} = 0\\ \Leftrightarrow – {x^{n + 2}} + 3{x^n} + {x^{n – 1}} – 2{x^6} = 0\\ \Leftrightarrow {x^6}\left( { – {x^{n – 4}} + 3{x^{n – 6}} + {x^{n – 7}} – 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} {x^6} = 0\\ – {x^{n – 4}} + 3{x^{n – 6}} + {x^{n – 7}} – 2 = 0 \end{array} \right.\\ \Leftrightarrow x = 0 \end{array}$

Câu d xem lại đề bài.

B2:

a) Ta có:

$\begin{array}{l} A = 2{x^4} – 61{x^3} – 95{x^2} – 31x + 64\\ = 2\left( {{x^4} – 32{x^3}} \right) + 3\left( {{x^3} – 32{x^2}} \right) – \left( {{x^2} – 32x} \right) – 63\left( {x – 32} \right) – 1952\\ = \left( {x – 32} \right)\left( {2{x^3} + 3{x^2} – x – 63} \right) – 1952 \end{array}$

Khi $x = 32 \Rightarrow A = – 1952$

b) Ta có:

$\begin{array}{l} B = {x^5} – 2013{x^4} + 2015{x^3} – 4032{x^2}\\ = \left( {{x^5} – 2012{x^4}} \right) – \left( {{x^4} – 2012{x^3}} \right) + 3\left( {{x^3} – 2012{x^2}} \right) + 2004{x^2}\\ = \left( {x – 2012} \right)\left( {{x^4} – {x^3} + 3{x^2}} \right) + 2004{x^2} \end{array}$

Khi $x = 2012 \Rightarrow B = {2004.2012^2}$