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Bài làm
a) x2 - 10x + 26
Thay x = 105 vào biểu thức trên, ta được:
1052 - 10 . 105 + 26
= 11025 - 1050 + 25
= 9975 + 25
= 10000
b) x2 + 0,2x + 0,01
Thay x = 0,9 vào biểu thức trên, ta được
0,92 + 0,2 . 0,9 + 0,01
= 0,81 + 0,18 + 0,01
= 1
c) 2( a - 5 )( a + 1 ) - ( a - 5 )2 + 36
= ( a - 5 )( 2a + 2 - a + 5 ) + 36
= ( a - 5 )( a + 7 ) + 36
Thay a = 99, ta được:
( 99 - 5 )( 99 + 7 ) + 36
= 94 + 106 + 36
= 236
a, Đặt A = \(x^2-10x+26\)
= \(x^2-10x+25+1\)
=\(x^2-2.x.5+5^2+1\)
=\(\left(x-5\right)^2+1\)
tại x = 105 thay vào A ta được :
A = \(\left(105-5\right)^2+1\)= 10000+1=10001
b Đặt B = \(x^2+0,2x+0,01\)
= \(x^2+2.x.0,1+0,1^2\)
= \(\left(x+0,1\right)^2\)
Tại x = 0,9 thay vào B ta được :
B = \(\left(0,9+0,1\right)^2=1^2=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{•}A=x^2-6x+10=\left(x-3\right)^2+1\\ A=\left(103-3\right)^2+1=10001\\ \text{•}B=x^2+0,2x+1,01\\ B=\left(1,01\right)^2+0,2.1,01+1,01=2,2321\\ \text{•}C=x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\\ C=\left(3,5-3,25.2\right)\left(3,5+3,25.2\right)=-30\)
\(A=x^2-6x+10\)
\(=x^2-2.x.3+3^2+1\)
\(=\left(x-3\right)^2+1\)
tại x=103, ta có:
\(\left(103-3\right)^2+1=10001\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128
b: \(B=\left(2x-4\right)^2+2\cdot\left(2x-4\right)\left(x+1\right)+\left(x+1\right)^2\)
=(2x-4+x+1)^2
=(3x-3)^2
Khi x=-1/2 thì B=(-3/2-3)^2=(-9/2)^2=81/4
c: \(C=x^2\left(5-4\right)+y^2\left(4-6\right)+z^2\left(6+4\right)\)
=x^2-2y^2+10z^2
=6^2-2*5^2+10*4^2
=146
d: x=9 thì x+1=10
\(D=x^{2017}-x^{2016}\left(x+1\right)+x^{2015}\left(x+1\right)-...-x^2\left(x+1\right)+x\left(x+1\right)-\left(x+1\right)\)
=x^2017-x^2017+x^2016+...-x^3-x^2+x^2+x-x-1
=-1
a: A=3(x^2-y^2)-2(x-y)^2
=3(x+y)(x-y)-2(x-y)^2
=(x-y)(3x+3y-2x+2y)
=(x-y)(x+5y)
=(4+4)(4-5*4)
=8*(-16)=-128
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=x^5-15x^4+16x^3-29x^2+13x\)
\(=x^5-14x^4-x^4+14x^3+2x^3-28x^2-x^2+14x-x+14-14\)
\(=x^4\left(x-14\right)-x^3\left(x-14\right)+2x^2\left(x-14\right)-x\left(x-14\right)-\left(x-14\right)-14\)
\(=\left(x^4-x^3+2x^2-x-1\right)\left(x-14\right)-14\)
Thay x = 14 => B = -14
Vậy...
phần còn lại tách ra làm tương tự nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
x=9
=>x+1=10
\(A=x^{10}-10x^9+10x^8-...+10x^2-10x+1\)
\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+1\)
\(=x^{10}-x^{10}-x^9+x^8+...+x^3+x^2-x^2-x+1\)
=-x+1
=-9+1=-8
![](https://rs.olm.vn/images/avt/0.png?1311)
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a) Ta có : \(x=31\Rightarrow30=x-1\)
Thay vào biểu thức ta được:
\(A=x^3-\left(x-1\right).x^2-x^2+1=x^3-x^3+x^2-x^2+1=1\)
b) Ta có: \(x=9\Rightarrow x+1=10\)
Thay vào biểu thức ta được
\(B=x^{14}-\left(x+1\right).x^{13}+\left(x+1\right).x^{12}-\left(x+1\right).x^{11}+.....+x^2.\left(x+1\right)=\left(x+1\right).x+\left(x+1\right)\)
\(\Leftrightarrow B=x^{14}-x^{14}-x^{13}+x^{13}+....+x^3+x^2=x^2+2x+1\)
\(\Leftrightarrow B=x^2-x^2-2x-1=-2.9-1=-19\)
\(x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+..+\left(x+1\right)x^2-\left(x+1\right)x+x+1\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+...+x^3+x^2-x^2-x+x+1\)
\(=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: x = 9 => x - 9 = 0
\(Q\left(x\right)=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
\(=x^{14}-9x^{13}-x^{13}+9x^{12}+x^{12}-9x^{11}+...-x^3+9x^2+x^2-9x-x+9+1\)
\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+...-x^2\left(x-9\right)+x\left(x-9\right)-\left(x-9\right)+1\)
\(=0+1=1\)
\(A=6xy\left(xy-y^2\right)-8x^2.\left(x-y^2\right)+5y^2\left(x^2-xy\right)\)
\(A=6x^2y^2-6xy^3-8x^3+8x^2y^2+5y^2x^2-5xy^3\)
\(A=19x^2y^2-11xy^3-8x^3\)
Tại x=1/2, y=2
\(A=19.\frac{1}{4}.2^2-11.\frac{1}{2}.2^3-8\left(\frac{1}{2}\right)^3=19-44-1=-26\)