Bạn chú thích hơi quá lố :) 

Ta có :( 5x - 3y + 4z ) . ( 5x - 3y - 4z ) \(=\left(5x-3y\right)^2-16z^2\)

\(=25x^2-30xy+9y^2-16z^2\)

Mà x^2=y^2 + z^2 nên ( 5x - 3y + 4z ) . ( 5x - 3y - 4z )\(=25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)

\(=9x^2-30xy+25y^2=\left(3x-5y\right)^2\)

Học tốt !

\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y^2\right)\)

\(\Leftrightarrow\left(5x-3y\right)^2-16z^2-\left(3x-5y\right)^2=0\)

\(\Leftrightarrow\left(5x-3y-3x+5y\right)\left(5x-3y+3x-5y\right)-16z^2=0\)

\(\Leftrightarrow16x^2=16y^2+16z^2\)(luôn đúng)

 \(a^2-3ab+2c=0\)

\(\Leftrightarrow\left(x+y\right)^2-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)=0\)

\(\Leftrightarrow x^2+y^2+2xy-3x^3-3xy^2-3yx^2-3y^3+2y^3+2x^3=0\)

\(\Leftrightarrow\left(x+y\right)^2-\left(x+y\right)^3=0\)

Hình như đề hơi sai nếu là a^3 thì ms đúng chứ

mk chỉ góp ý thui tại làm mãi ko ra

hhihihihi

ý bn là chia hết cho 31 hả ?

đây là câu chia hết cho 31 nhé , em ghi nhầm

\(A=\left(2+2^2+2^3+2^4+2^5\right)+\)\(+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\)\(+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(A=2.\left(2^2+2^3+2^4+2^5\right)+...+\)\(2^{96}.\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(A=2.31+...+2^{96}.31⋮31\)

\(\Rightarrow A⋮31\)

(23+24+25+26)-(13+14+15+16)
= 23+24+25+26-23-24-25-26
=(23-23)+(24-14)+(25-25)+(26-26)
=10+10+10+10=40

=5+5+5+...+5+5
=5x40=200

(23+24+25+26)-(13+14+15+16)
=5+5+5+...+5+5
=5x40=200

a) Ta có: P(x) = 2x⁵ + 2 - 6x² - 3x³ + 4x² - 2x + x³ + 4x⁵

                       = (2x⁵ + 4x⁵) + 2 - (6x² - 4x²) - (3x³ - x³) - 2x

                      =       6x⁵ + 2 - 2x² - 2x³ - 2x

b) P(x) = 6x⁵ - 2x³ - 2x² - 2x + 2

c) Bậc của P(x) là 5

a,\(P\left(x\right)=2^5+2-6x^2-3x^3+4x^2-2x+x^3+4x^5\)

   \(P\left(x\right)=(2x^5+4x^5)+\left(-6x^2+4x^2\right)+\left(-3x^3+x^3\right)+2-2x\)

   \(P\left(x\right)=6x^5-2x^2-2x^3+2-2x\)

b, \(P\left(x\right)=6x^5-2x^3-2x^2-2x+2\)

c, Bậc của P(x)=5

a) \(\left(2x+1\right)^2-4\left(x+2\right)^2=12\)

\(\Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)=12\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16-12=0\)

\(\Leftrightarrow-12x-27=0\)

\(\Leftrightarrow x=\frac{-9}{4}\)

b) xem lại đề

c) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x-3\right)\left(3-x\right)=1\)

\(\Leftrightarrow x^3-27-x\left(x-3\right)^2=1\)

\(\Leftrightarrow x^3-27-x\left(x^2-6x+9\right)-1=0\)

\(\Leftrightarrow x^3-28-x^3+6x^2-9x=0\)

\(\Leftrightarrow6x^2-9x-28=0\)

\(\Leftrightarrow6\left(x^2-\frac{3}{2}x-\frac{14}{3}\right)=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}-\frac{251}{48}=0\)

\(\Leftrightarrow\left(x-\frac{3}{4}\right)^2=\frac{251}{48}=\left(\pm\sqrt{\frac{251}{48}}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{4}=\sqrt{\frac{251}{48}}=\frac{\sqrt{753}}{12}\\x-\frac{3}{4}=-\sqrt{\frac{251}{48}}=\frac{-\sqrt{753}}{12}\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{\pm\sqrt{753}}{12}+\frac{3}{4}=\frac{9\pm\sqrt{753}}{12}\)

d) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)

\(\Leftrightarrow12x+15=0\)

\(\Leftrightarrow x=\frac{-5}{4}\)

Theo giả thiết:

\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Dễ thấy \(VT\ge0\forall a;b;c\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)(đpcm)

Cách 1:

Ta có: x² – x – 12 = x² – x – 16 + 4

= (x² – 16) – (x – 4)

= (x – 4).(x + 4) – (x – 4)

= (x – 4).(x + 4 – 1)

= (x – 4).(x + 3)

Cách 2:

x² – x – 12 = x² – x – 9 – 3

= (x² – 9) – (x + 3)

= (x – 3).(x + 3) – (x + 3)

= (x + 3).(x – 4)

Cách 3:

x² – x – 12 = x² – 4x + 3x – 12

= (x² – 4x) + (3x – 12)

= x.(x – 4) + 3.(x – 4)

= (x – 4).(x + 3).

Chúc bạn hk tốt nha.Nhớ cho mik nhé bạn

#include <bits/stdc++.h>

using namespace std;

long long n,i,s;

int main()

{

cin>>n;

if (n%2==0)

{

s=1;

for (i=1; i<=n; i++)

if (i%2==0) s=s*i;

cout<<s;

}

else 

{

s=1;

for (i=1; i<=n; i++)

if (i%2==1) s=s*i;

cout<<s;

}

return 0;

}

a) \(6x^2-11x+3\)

\(=6x^2-9x-2x+3\)

\(=3x\left(2x-3\right)-\left(2x-3\right)\)

\(=\left(3x-1\right)\left(2x-3\right)\)

b) \(2x^2+3x-27\)

\(=2x^2-6x+9x-27\)

\(=2x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(2x+9\right)\left(x-3\right)\)

f) \(x^2+4xy+3y^2\)

\(=x^2+xy+3xy+3y^2\)

\(=x\left(x+y\right)+3y\left(x+y\right)\)

\(=\left(3y+x\right)\left(x+y\right)\)

\(x^2-x+1-m=0\left(1\right)\\ \text{PT có 2 nghiệm }x_1,x_2\\ \Leftrightarrow\Delta=1-4\left(1-m\right)\ge0\\ \Leftrightarrow4m-3\ge0\Leftrightarrow m\ge\dfrac{3}{4}\\ \text{Vi-ét: }\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=1-m\end{matrix}\right.\\ \text{Ta có }5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\\ \Leftrightarrow5\cdot\dfrac{x_1+x_2}{x_1x_2}-x_1x_2+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m-1+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m+3=0\\ \Leftrightarrow5+\left(1-m\right)\left(m+3\right)=0\\ \Leftrightarrow m^2+2m-8=0\\ \Leftrightarrow m^2-2m+4m-8=0\\ \Leftrightarrow\left(m-2\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\left(n\right)\\m=-4\left(l\right)\end{matrix}\right.\)

Vậy $m=2$