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\(A=x^{10}-14x^9-x^9+14x^8+x^8-14x^7-x^7...-x+14+1\)
\(A=x^9\left(x-14\right)-x^8\left(x-14\right)+x^7\left(x-14\right)-...-x\left(x-14\right)+1\)
\(A=1\) (Do x=14)
a: \(=\dfrac{5\cdot3x+4x-3-2\left(15x-1\right)}{10}=\dfrac{19x-3-30x+2}{10}=\dfrac{-11x-1}{10}\)
b: \(\Leftrightarrow x\left(x-4\right)+\left(x+1\right)^2=2x\left(x+1\right)\)
\(\Leftrightarrow x^2-4x+x^2+2x+1=2x^2+2x\)
=>-2x+1=2x
=>-4x=-1
hay x=1/4(nhận)
\(x^3-15x^2+75x-125\)
\(=x^3-3.5x^2+3.5^2.x-5^3\)
= \(\left(x+5\right)^3\)
Thay x=35 và biểu thức trên ta có :
\(\left(35+5\right)^3\)= 403= 64000
a) \(x^3+15x^2+75x=-125\)
\(\Leftrightarrow x^3+15x^2+75x+125=0\)
\(\Leftrightarrow x^3+125+15x^2+75x=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+5x+25\right)+15x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2+20x+25\right)=0\)
\(TH1:x+5=0\Leftrightarrow x=-5\)
\(TH2:x^2+20x+25=0\)
\(\Leftrightarrow\left(x+10\right)^2=75\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{75}-10\\x=-\sqrt{75}-10\end{cases}}\)
ta có
a. (5x-7)(x-9)-(-x+3)(-5x+2)= 2x(x-4)-(x-1)(2x+3)
\(\Leftrightarrow5x^2-52x+63-\left(5x^2-17x+6\right)=2x^2-8x-\left(2x^2+x-3\right)\)
\(\Leftrightarrow-35x+57=-9x+3\Leftrightarrow26x=54\Leftrightarrow x=\frac{27}{13}\)
b. (x-3)(-x+10)+(x-8)(x+3)= (5x^2-1)(x+3)-5x^3-15x^2
\(\Leftrightarrow-x^2+13x-30+x^2-5x-24=5x^3+15x^2-x-3-5x^3-15x^2\)
\(\Leftrightarrow8x-54=-x-3\Leftrightarrow9x=51\Leftrightarrow x=\frac{17}{3}\)
a: \(\Leftrightarrow5x^2-45x-7x+63-\left(5x-2\right)\left(x-3\right)=2x^2-8x-2x^2-3x+2x+3\)
\(\Leftrightarrow5x^2-52x+63-\left(5x-2\right)\left(x-3\right)=-9x+3\)
\(\Leftrightarrow5x^2-52x+63-5x^2+15x+2x-6=-9x+3\)
=>-37x+57=-9x+3
=>28x=-54
hay x=-27/14
b: \(\Leftrightarrow-x^2+19x+3x-30+x^2-5x-24=\left(5x^2-1\right)\left(x+3\right)-5x^3-15x^2\)
\(\Leftrightarrow17x-54=5x^3+15x^2-x-3-5x^3-15x^2\)
=>18x=51
hay x=17/6
e) Ta có: \(E=\left(3x+2\right)\left(3x-5\right)\left(x-1\right)\left(9x+10\right)+24x^2\)
\(=\left(9x^2-15x+6x-10\right)\left(9x^2+10x-9x-10\right)+24x^2\)
\(=\left(9x^2-10-9x\right)\left(9x^2-10+x\right)+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)-9x^2+24x^2\)
\(=\left(9x^2-10\right)^2-8x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)^2-3x\left(9x^2-10\right)-5x\left(9x^2-10\right)+15x^2\)
\(=\left(9x^2-10\right)\left(9x^2-3x-10\right)-5x\left(9x^2-10-3x\right)\)
\(=\left(9x^2-3x-10\right)\left(9x^2-5x-10\right)\)
B2
( a3 + a2b + ab2 + b3 ).( a - b ) = a4 - b4
[( a3 + b3 + ab.( a + b )].( a - b ) = a4 - b4
[( a + b ).( a2 - ab + b2 ) + ab.( a + b )].( a - b ) = a4 - b4
( a + b ).( a2 - ab + b2 + ab ).( a - b ) = a4 - b4
( a + b ).( a2 + b2 ).( a - b ) = a4 - b4
( a2 - b2 ).( a2 + b2 ) = a4 - b4
a4 - b4 = a4 - b4 ( đpcm )
Ta có x=14 suy ra x+1=15
Do đó thay x+1 vào H(x), ta được:
H(14)= x10 - (x+1)x9 +(x+1)x8-(x+1)x7+...+ (x+1)x2 - (x+1)x + x+1
H(14)= x^10 - x^10 -x^9 +x^8- x^8-x^7 +....+ x^3 +x^2 -x^2-x+x +1
Hay H(14)=1
Đặt Q(x) = x9 - x8 + x7 - ... + x - 1 thì (x + 1) * Q(x) = (x + 1) * (x9 - x8 + x7 - ... + x - 1) = x10 - 1 \(\Rightarrow\left(14+1\right)\cdot Q\left(14\right)=14^{10}-1\)
Dễ thấy: H(x) = x10 - 15* Q(x) \(\Rightarrow H\left(14\right)=14^{10}-\left(14^{10}-1\right)=1\)
Xin cậu !
A=-103+15.-102+75+125
A=-1000+15.100+75+125
A=-1000+1500+75+125
A=500+200
A=700
Với x=-10 ta có:
A=(-10)^3+15.(-10)^2+75+125
A=-1000+15.100+75+125
A=-1000+1500+75+125
A=500+75+125
A=575+125
A=700
Vậy với x=-10 ta có A=700