x³-x²+2x+7 chia x²+1 dư x+8

=>x³-x²+2x+7 chia hết cho x²+1<=>x+8=0<=>x=-8(thỏa mãn x thuộc Z)

Vậy để x³-x²+2x+7 chia hết cho x²+1 thì x=-8

\(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)

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

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

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

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

\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)\left(2x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\)\(x=-\frac{3}{2}\) hoặc \(x=1\) hoặc \(x=\frac{1}{2}\) hoặc \(x=-1\)

\(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)

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

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

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

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

\(\Leftrightarrow\left[2x\left(x+1\right)-\left(x+1\right)\right]\left[2x\left(x-1\right)+3\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-1\right)\left(2x+3\right)\left(x-1\right)=0\)

Nếu x + 1 = 0 thì x = -1

Hoặc 2x - 1= 0 thì x = 1/2

Hoặc 2x + 3 = 0 thì x = -3/2

Hoặc x - 1 = 0 thì x = 1

Vậy ....

ồ cuk dễ nhỉ

Nếu các bn thích thì ...........

cứ cho NTN này nhé !

TA có \(\left(a+b+c\right)^2=0\Rightarrow ab+bc+ca=-\frac{1}{2}\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)

=> \(a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)

Mà \(\left(a^2+b^2+c^2\right)^2=1\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)

=> \(a^4+b^4+c^4=\frac{1}{2}\)

^_^

Ta có: a+b+c=0 <=> (a+b+c)²=0 <=> a²+b²+c²+ 2( ab+ac+bc)=0 <=> 2(ab+ac+bc)= -1 ( vì a²+b²+c²=1) <=> ab+ac+bc= -1/2 

=> (ab+ac+bc)²= 1/4 <=> a²b²+a²c²+b²c²+2abc(a+b+c)= 1/4 <=> 2(a²b²+a²c²+b²c²)= 1/2 ( vì a+b+c=0) (*)

Lại có: a²+b²+c²=1 <=> (a²+b²+c²)²=1 <=> a⁴+b⁴+c⁴+2(a²b²+a²c²+b²c²)=1 <=> a⁴+b⁴+c⁴= 1/2 ( vì (*))

Vậy,...

\(x^2+10x+26+y^2+2y\)

\(=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

\(\left(x+y+4\right)\left(x+y-4\right)\)

\(=\left(x+y\right)^2-16\)

\(=x^2+y^2+2xy-16\)

a, =(x^2 +10x+25) +(y^2 +2y+1)

    = (x+5)^2 +(y+1)^2

b, =(x+y)^2 -4^2

    = x^2 + 2xy+ y^2 -16

Ta có: góc B- góc A=20⁰ <=> Góc B= góc A+20⁰ (1) ; góc C= 3 góc A ( giả thiết) (2) ; góc D- góc C=20⁰ <=> góc D= 3 góc A+20⁰ (theo(2))

 Mà : góc A+ góc B+ góc C+ góc D=360⁰ (*). Thay (1);(2);(3) vào (*), ta được: Góc A+ góc A+20⁰+3 góc A+3 góc A+20⁰=360⁰

<=> Góc A= 40⁰ => Các góc còn lại

Gọi số đo góc A  là  x  

thì số đo góc B là:  x + 20

     số đo góc C là:  3x    =>   số đo góc D là:   3x + 20

Ta có:     \(x+\left(x+20\right)+3x+\left(3x+20\right)=180\)

        \(\Leftrightarrow\)\(8x=140\)

       \(\Leftrightarrow\)\(x=17,5\)

Vậy góc A = 17,5⁰

       góc B = 17,5⁰ + 20⁰ = 37,5⁰

       góc C = 17,5 . 3 = 52,5⁰

       góc D = 52,5⁰ + 20⁰ = 72,5⁰

Đặt \(A=1+2+2^2+...+2^{32}\)

\(\Rightarrow2A=2+2^2+2^3+...+2^{33}\)

\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{33}\right)-\left(1+2+2^2+...+2^{32}\right)\)

\(\Rightarrow A=2^{33}-1\)

\(\Rightarrow n=33\)

Vậy n = 33

_Chúc bạn học tốt_

1+2+2^2+2^3+2^4+...+2^32=2^n-1 (1)

=>2+2^2+2^3+...+2^33=2^(n+1)-2 (2)

=>trừ (2) cho (1) ta có : 2^33-1=(2-1)*(2^n-1)

=>2^33-1=2^n-1

=>n=33

vậy n=33

k cho mình nha

6)   \(9x^2+6xy+y^2=\left(3x+y\right)^2\)

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

8)   \(x^2-2xy+y^2-16=\left(x-y\right)^2-16=\left(x-y-4\right)\left(x-y+4\right)\)

9)   \(4x^2-y^2+4x+1=\left(2x+1\right)^2-y^2=\left(2x-y+1\right)\left(2x+y+1\right)\)

10)  \(x^3-x+y^3-y=\left(x+y\right)\left(x^2-xy+y^2+1\right)\)

6) = (3x)² + 2.(3x)y +y² = (3x + y)²

7) = (x-y)(x+y)- 3(x+y) = (x+y)(x-y-3)

8) = (x-y)²  - 4² = (x-y-4)(x-y+4)

9) = ( 4x² + 4x +1 ) - y² = (2x+1)² - y^2 = (2x+1-y)(2x+1+y)

10) =(x³+y³) - (x+y) = (x+y)(x²+xy+y²) - (x+y) = (x+y)(x²+xy+y²-1)

k mk đi nha

a)   \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)

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

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

d)  \(64+16y+y^2=\left(8+y\right)^2\)

c) mk chỉnh đề:

\(16-\left(x-3\right)^2=\left(4+x-3\right)\left(4-x+3\right)=\left(x+1\right)\left(7-x\right)\)

ồ cuk dễ nhỉ

Nếu các bn thích thì ...........

cứ cho NTN này nhé !

\(x^2-2xy+y^2-z^2+2zt-t^2\)

\(=\left(x^2-2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)

\(=\left(x-y\right)^2-\left(z-t\right)^2\)

\(=\left(x-y+z-t\right)\left(x-y-z+t\right)\)

hok tốt!

\(x^2-2xy+y^2-z^2+2zt-t^2\)

\(=\left(x-y\right)^2-\left(z-t\right)^2\)

\(=\left(x-y+z-t\right)\left(x-y-z+t\right)\)