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\(A=10x^2-7x-5=\left(10x^2-15x\right)+8x-12+7=5x\left(2x-3\right)+4\left(2x-3\right)+7\)
\(A⋮B\Leftrightarrow7⋮2x+3\)
Rồi xét từng ước và tìm x
Ta có : \(h\left(x\right)=10x^2-7x-5\)
\(=10x^2-15x+8x-12+7\)
\(=5x\left(2x-3\right)+4\left(2x-3\right)+7\)
\(=\left(5x+4\right)\left(2x-3\right)+7\)
Để \(h\left(x\right)⋮k\left(x\right)\)
\(\Leftrightarrow\left(5x+4\right)\left(2x-3\right)+7⋮2x-3\)
\(\Leftrightarrow7⋮2x-3\)
Do \(x\in Z\Rightarrow2x-3\in Z\)
\(\Rightarrow2x-3\in\left\{\pm1;\pm7\right\}\)
Ta có bảng sau :
| \(2x-3\) | \(1\) | \(-1\) | \(7\) | \(-7\) |
| \(x\) | \(2\) | \(1\) | \(5\) | \(-2\) |
Vậy \(x\in\left\{\pm2;1;5\right\}\)
1
a) x^2+2x-5 b) x^2+x+7 9 (dư 8)
2
x=2; x = -(3*căn bậc hai(7)*i+1)/2;x = (3*căn bậc hai(7)*i-1)/2;
3
a=2
a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)
Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)
Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)
\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)
\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4
Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4
a) A= \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\)
\(ĐK:3x^2-7x+2\ne0\)
\(\Leftrightarrow\orbr{\begin{cases}x\ne\frac{1}{3}\\x\ne2\end{cases}\left(^∗\right)}\)
=> 3x2 + 5x + 2 =0
<=> 3x2 + 3x + 2x +2 = 0
<=> 3x .( x + 1 ) + 2 .( x + 1 ) =0
<=> ( x + 1 )(3x + 2 ) =0
<=> \(\orbr{\begin{cases}x+1=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{-2}{3}\left(t/m\left(^∗\right)\right)\end{cases}}}\)
Vậy x = -2/3
b) \(B=\frac{2x^2+10x+12}{x^3-4x}=0\left(ĐK:x\ne0;x^2\ne4\Leftrightarrow x\ne0;x\ne\pm2\right)\)
<=> 2x2+ 10x + 12 = 0
<=> x2 + 5x+ 6 =0
<=> ( x + 2 ) ( x + 3 ) =0\(\Leftrightarrow\orbr{\begin{cases}x=-2\left(L\right)\\x=-3\left(t/m\right)\end{cases}}\)
Vậy x = -3
c)\(C=\frac{x^3+x^2-x-1}{x^3+2x-5}=0\) \(ĐK:x^3+2x-5\ne0\left(^∗\right)\)
<=> x3 + x2 -x -1 =0
<=> ( x - 1 )(x2 + 2x + 1 )
<=> ( x-1 ) (x+1)2 = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(t/m\left(^∗\right)\right)\\x=-1\left(t/m\left(^∗\right)\right)\end{cases}}}\)
Vậy x = { 1 ; -1 }
a) A = \(\frac{3x^2+5x-2}{3x^2-7x+2}=0\) (ĐKXĐ: x khác 1/3, x khác 2)
<=> 3x^2 + 5x - 2 = 0
<=> (3x - 1)(x + 2) = 0
<=> 3x - 1 = 0 hoặc x + 2 = 0
<=> 3x = 1 hoặc x = -2
<=> x = 1/3 (ktm) hoặc x = -2 (tm)
=> x = -2
b) B = \(\frac{2x^2+10x+12}{x^3-4x}=0\) (ĐKXĐ: x khác 0, x khác +-2)
<=> \(\frac{2\left(x^2+5x+6\right)}{x\left(x^2-4\right)}=0\)
<=> \(\frac{2\left(x+2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
<=> \(\frac{2\left(x+3\right)}{x\left(x-2\right)}=0\)
<=> 2(x + 3) = 0
<=> x + 3 = 0
<=> x = -3
c) C = \(\frac{x^3+x^2-x-1}{x^3+2x-5}=0\) (ĐKXĐ: x khác x^3 + 2x - 5)
<=> \(\frac{x^2\left(x+1\right)-\left(x+1\right)}{x^3+2x-5}=0\)
<=> \(\frac{\left(x+1\right)\left(x^2-1\right)}{x^3+2x-5}=0\)
<=> \(\frac{\left(x+1\right)\left(x-1\right)\left(x+1\right)}{x^3+2x-5}=0\)
<=> (x + 1)(x - 1) = 0
<=> x + 1 = 0 hoặc x - 1 = 0
<=> x = -1 hoặc x = 1
Bài 2:
a) Vì x = 79 => x + 1 = 80
\(P\left(x\right)=x^7-80x^6+80x^5-80x^4+.....+80x+15\)
\(\Rightarrow P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+.....+\left(x+1\right)x+15\)
\(=x^7-x^7-x^6+x^6+x^5-x^5-x^4+....+x^2+x+15\)
\(=x+15\)
Thay x = 79 vào đa thức ta được:
79 + 15 = 94
b) Vì x = 9 => x + 1 = 10
\(Q\left(x\right)=x^{14}-10x^{13}+10x^{12}-10x^{11}+.....+10x^2-10x+10\)
\(\Rightarrow Q\left(x\right)=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-\left(x+1\right)x^{11}+....+\left(x+1\right)x^2-\left(x+1\right)x+10\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-x^{12}-x^{11}+....+x^3+x^2-x^2-x+10\)
\(=-x+10\)
\(=-9+10=1\)
P/s: Ko chắc nhé!
Bài 1:
a/ \(\left(2x-1\right)\left(x^2-x+1\right)-2x^3+3x^2=2\)
\(\Rightarrow2x\left(x^2-x+1\right)-1\left(x^2-x+1\right)-2x^3+3x^2=2\)
\(\Rightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2=2\)
\(\Rightarrow3x-1=2\)
\(\Rightarrow3x=2+1=3\)
\(\Rightarrow x=3:3=1\)
b/ \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Rightarrow x\left(x^2+2x+4\right)+1\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Rightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Rightarrow6x+20=0\)
\(\Rightarrow6x=0-20=-20\)
\(\Rightarrow x=-\frac{20}{6}=-\frac{10}{3}\)
c/ \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow\left[x\left(x+2\right)+1\left(x+2\right)\right]\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow\left(x^2+2x+x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow x^2\left(x+5\right)+3x\left(x+5\right)+2\left(x+5\right)-x^3-8x^2=27\)
\(\Rightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2=27\)
\(\Rightarrow17x+10=27\)
\(\Rightarrow17x=27-10=17\)
\(\Rightarrow x=17:17=1\)