b/ \(4\left(x-1\right)\left(x+1\right)-5x\left(x-2\right)+x^2\)

= \(4\left(x^2-1\right)-5x^2+10x+x^2\)

= \(4x^2-4-5x^2+10x+x^2\)

= \(10x-4\)

= \(2\left(5x-2\right)\)

c/ \(\left(3-2x\right)\left(x-2\right)+4\left(x-1\right)\left(x-3\right)-2\left(x-2\right)\left(x+2\right)\)

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

= \(3x-6-2x^2+4x+4x^2-16x-16-8x^2-18\)

= \(-9x-12\)

= \(-3\left(3x+4\right)\)

ABCD là hình thang cân (gt) nên AB song song với CD,AD=BC=6cm và góc C=góc ADC

DB la tia p/g của góc ADC(gt) nên góc ADB=góc BDC= 1/2 góc ADC =1/2 góc C

AB song song với CD (cmt) suy ra: góc ABD=góc BDC

Tam giác ABD có: góc ABD=góc ADB(=góc BDC)

Do đó tam giác ABD cân tại A (DHNB) suy ra: AB=AD=6cm

Tam giác DBC vuông tại B nên góc BDC+góc C=90 độ

Hay 1/2 góc C+ góc C=90 độ

3/2 góc C =90 độ

C=60 độ.Sau đó tính được góc BDC=30 độ

Tam giác BDC vuông tại B có góc BDC=30 độ vì thế BC=1/2 DC

Do đó:DC=2BC=2x6=12(cm)

Chu vi hình thang ABCD là:

                             AB+AD+BC+CD=6+6+6+12=30(cm)

Vậy chu vi hình thang ABCD là 30 cm

P/s chu vi hình thang 30 cm :)))))

Ta có 10=9+1=x+1(Vì x=9)

=>B= x¹⁴-(x+1)x¹³+(x+1)x¹²-(x+1)x¹¹+.........-(x+1)x+10

=>B= x¹⁴-x¹⁴-x¹³+x¹³+x¹²-x¹²-x¹¹+.....-x²-x+10

=>B=-x+10

Thay x=9, ta có

B=-9+10=1

\(B=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)

\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+...+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)

\(=1\)

  Ta có:x⁵-15x⁴+16x³-29x³+13x=x⁵-15x⁴-13x³+13x

  Thay x=4 vào bt, ta có:

    4⁵-15.4⁴-13.4³+13.4

=1024-3840-832+52

=-3596

3596 nh bn

học tốt

a) Ta có: \(A=x^5-15x^4+16x^3-29x^2+13x\)

\(=\left(x^5-14x^4\right)-\left(x^4-14x^3\right)+\left(2x^3-28x^2\right)-\left(x^2-14x\right)-x\)

\(=x^4\left(x-14\right)-x^3\left(x-14\right)+2x^2\left(x-14\right)-x\left(x-14\right)-x\)

\(=\left(x-14\right)\left(x^4-x^3+2x^2-x\right)-x\)(thay x = 14)

\(=-x=-14\)

Vậy A = -14.

b) Ta có: \(B=x^{14}-10x^3+10x^{12}-10x^{11}+...+10x^2-10x+10\) tại x = 9.

\(\cdot x=9\Rightarrow10=x+1\)

\(\Rightarrow B=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+...+\left(x+1\right)x^2-\left(x+1\right)x+10\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{13}-x^{12}+...+x^3+x^2-x^2-x+10\)

\(=-x-10=-9-10=-19.\)

Vậy B = -19.

a) Ta có:

\(A=x^5-15x^4+16x^3-29x^2+13x\)

\(=\left(x^5-14x^4\right)-\left(x^4-14x^3\right)+\left(2x^3-28x^2\right)-\left(x^2-14x\right)-x\)

\(=x^4\left(x-14\right)-x^3\left(x-14\right)+2x^2\left(x-14\right)-x\left(x-14\right)-x\)

\(=\left(x-14\right)\left(x^4-x^3+2x^2-x\right)-x\)(thay \(x=14\))

\(=-x=-14\)

Vậy \(A=-14\)

b) Ta có:

\(B=x^{14}-10x^3+10x^{12}-10x^{11}+...+10x^2-10x+10\)tại \(x=9\)

\(x=9\Rightarrow10=x+1\)

\(\Rightarrow B=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+...+\left(x+1\right)x^2-\left(x+1\right)x+10\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{13}-x^{12}+...+x^3+x^2-x^2-x+10\)

\(=-x-10=-9-10=-19\)

Vậy \(B=-19\)

a) Ta có:

(x+y+z)(x-y-z) = x^2 -xy -xz +yx- y^2 -yz+zx -zy -z^2

=x^2 - y^2 - 2yz - z^2.

b) Ta có: (x-y+z)(x+y+z) = x^2 +xy+xz -yx-y^2 -yz +zx+zy +z^2

=x^2 +2xz- y^2 +z^2.

c) Ta có: -16 + (x-3)^2 = -16 + ( x^2-6x+9)

= -16 + x^2 - 6x + 9

= x^2 - 6x - 7.

\(a,\left(x+y+z\right)\left(x-y-z\right)\)

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

\(=x^2-xy-xz+xy-y^2-yz+xz-yz-z^2\)

\(=x^2-y^2-2yz-z^2\)

\(b,\left(x-y+z\right)\left(x+y+z\right)\)

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

\(=x^2+xy+xz-xy-y^2-yz+xz+yz+z^2\)

\(=x^2+2xz-y^2+z^2\)

\(c,-16+\left(x-3\right)^2\)

\(=-16+x^2-6x+9\)

\(=x^2-6x-7\)

\(a,x^2-2x=24\)

\(x^2-2x-24=0\)

\(x^2-2x+1-25=0\)

\(\left(x-1\right)^2=5^2=\left(-5\right)^2\)

\(x-1=5\)                     hoặc                           \(x-1=-5\)

\(\Rightarrow\hept{\begin{cases}x=6\\x=-4\end{cases}}\)

\(b,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(2x+255=0\)

\(2x=-255\)

\(x=-\frac{255}{2}\)

a/ \(x^2-2x=24\)

<=> \(x^2-2x+1-1=24\)

<=> \(\left(x-1\right)^2=25\)

<=> \(\orbr{\begin{cases}x-1=25\\x-1=-25\end{cases}}\)<=> \(\orbr{\begin{cases}x=26\\x=-24\end{cases}}\)

b/ \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

<=> \(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

<=> \(4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

<=> \(2x+255=0\)

<=> \(2x=-255\)

<=> \(x=-\frac{255}{2}\)

giai giup minh dc ko minh dang can gap

\(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\ge4>0\)

\(x^2+3x+6=x^2+3x+\frac{9}{4}+\frac{15}{4}=\left(x+\frac{3}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)