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PT
27 tháng 4 2021
P(x) = 9x4 + 5x3 + 8x2 - 15x3 - 4x2 - x4 + 15 - 7x2
= (9x4 - x4) + (5x3 - 15x3) + (8x2 - 4x2 - 7x2) + 15
= 8x4 - 10x3 - 3x2 + 15
Ta có: P(1) = 8. 14 - 10. 13 - 3. 12 + 15 = 8 - 10 - 3 + 15 = 10
P(0) = 8. 04 - 10. 03 - 3. 02 + 15 = 0 - 0 - 0 + 15 = 15
P(-1) = 8.(-1)4 - 10(-1)3 - 3(-1)2 + 15 = -8 - (-10) - (-3) + 15 = 20
19 tháng 8 2020
a,Cách 1 : \(x^2-10x+9=0\Leftrightarrow\left(x-1\right)\left(x-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=9\end{cases}}\)
Cách 2 : Dung p^2 nhẩm nghiệm p^2 bậc 2 vì : 1 - 10 + 9 = 0
\(\Leftrightarrow\orbr{\begin{cases}x_1=1\\x_2=\frac{c}{a}=9\end{cases}}\)
b, Cách 1 : \(8x^2-2x-15=0\Leftrightarrow\left(4x+5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=\frac{3}{2}\end{cases}}\)
Cách 2 : \(\Delta=\left(-2\right)^2-4.8.\left(-15\right)=484>0\)
Pp có 2 nghiệm phân biệt : \(x_1=\frac{-2-\sqrt{484}}{16};x_2=\frac{-2+\sqrt{484}}{16}\)
PN
20 tháng 8 2020
toán 9 à bạn ?
c,\(2x^2+8x-7=0\)
Ta có : \(\Delta=8^2-4.\left(-7\right).2=64+56=120\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-8+\sqrt{120}}{4}=-2+\frac{\sqrt{120}}{4}\\x=\frac{-8-\sqrt{120}}{4}=-2-\frac{\sqrt{120}}{4}\end{cases}}\)
d,\(3x^2-15x+3=0\)
Ta có : \(\Delta=\left(-15\right)^2-4.3.3=225-36=189\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15+\sqrt{189}}{6}\\x=\frac{15-\sqrt{189}}{6}\end{cases}}\)
e,\(16x^2-24x-4=0\Leftrightarrow4x^2-6x-1=0\)
Ta có : \(\Delta=\left(-6\right)^2-4.4.\left(-1\right)=36+16=52\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6+\sqrt{52}}{8}\\x=\frac{6-\sqrt{52}}{8}\end{cases}}\)
f, \(-5x^2+6x+3=0\)
Ta có : \(\Delta=6^2-4.3.\left(-5\right)=36+60=96\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-6+\sqrt{96}}{-10}\\x=\frac{-6-\sqrt{96}}{-10}\end{cases}}\)
i, \(6x^2-9x+40=0\)
Ta có : \(\Delta=\left(-9\right)^2-4.6.40=81-960=-879\)
do đen ta < 0 => vô nghiệm
18 tháng 8 2020
a)
Cách 1:
Ta có: \(x^2-10x+9=0\)
\(\Leftrightarrow x^2-x-9x+9=0\)
\(\Leftrightarrow x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)
Vậy: S={1;9}
Cách 2:
Ta có: \(x^2-10x+9=0\)
\(\Leftrightarrow x^2-10x+25-16=0\)
\(\Leftrightarrow\left(x-5\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy: S={9;1}
b)
Cách 1:
Ta có: \(8x^2-2x-15=0\)
\(\Leftrightarrow8x^2-12x+10x-15=0\)
\(\Leftrightarrow4x\left(2x-3\right)+5\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3}{2};\frac{-5}{4}\right\}\)
Cách 2:
Ta có: \(8x^2-2x-15=0\)
\(\Leftrightarrow8\left(x^2-\frac{1}{4}x-\frac{15}{8}\right)=0\)
\(\Leftrightarrow x^2-\frac{1}{4}x-\frac{15}{8}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{8}+\frac{1}{64}-\frac{121}{64}=0\)
\(\Leftrightarrow\left(x-\frac{1}{8}\right)^2=\frac{121}{64}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{8}=\frac{11}{8}\\x-\frac{1}{8}=-\frac{11}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{12}{8}=\frac{3}{2}\\x=\frac{-11+1}{8}=\frac{-10}{8}=\frac{-5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3}{2};\frac{-5}{4}\right\}\)
c) Ta có: \(2x^2+8x-7=0\)
\(\Leftrightarrow2\left(x^2+4x-\frac{7}{2}\right)=0\)
\(\Leftrightarrow x^2+4x+4-\frac{15}{2}=0\)
\(\Leftrightarrow\left(x+2\right)^2=\frac{15}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=\sqrt{\frac{15}{2}}\\x+2=-\sqrt{\frac{15}{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\frac{15}{2}}-2\\x=-\sqrt{\frac{15}{2}}-2\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{\frac{15}{2}}-2;-\sqrt{\frac{15}{2}}-2\right\}\)
d) Ta có: \(3x^2-15x+3=0\)
\(\Leftrightarrow3\left(x^2-5x+1\right)=0\)
\(\Leftrightarrow x^2-5x+1=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{21}{4}=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{21}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{5}{2}=\frac{\sqrt{21}}{2}\\x-\frac{5}{2}=-\frac{\sqrt{21}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{21}+5}{2}\\x=\frac{-\sqrt{21}+5}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{\sqrt{21}+5}{2};\frac{-\sqrt{21}+5}{2}\right\}\)
e) Ta có: \(16x^2-24x-4=0\)
\(\Leftrightarrow4\left(4x^2-6x-1\right)=0\)
\(\Leftrightarrow4x^2-6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-2\cdot2x\cdot\frac{3}{2}+\frac{9}{4}-\frac{13}{4}=0\)
\(\Leftrightarrow\left(2x-\frac{3}{2}\right)^2=\frac{13}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{3}{2}=\frac{\sqrt{13}}{2}\\2x-\frac{3}{2}=-\frac{\sqrt{13}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{3+\sqrt{13}}{2}\\2x=\frac{3-\sqrt{13}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{13}}{2}:2=\frac{3+\sqrt{13}}{4}\\x=\frac{3-\sqrt{13}}{2}:2=\frac{3-\sqrt{13}}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3+\sqrt{13}}{4};\frac{3-\sqrt{13}}{4}\right\}\)
f) Ta có: \(-5x^2+6x+3=0\)
\(\Leftrightarrow-5\left(x^2-\frac{6}{5}x-\frac{3}{5}\right)=0\)
\(\Leftrightarrow x^2-\frac{6}{5}x-\frac{3}{5}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{5}+\frac{9}{25}-\frac{24}{25}=0\)
\(\Leftrightarrow\left(x-\frac{3}{5}\right)^2=\frac{24}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{5}=\frac{2\sqrt{6}}{5}\\x-\frac{3}{5}=\frac{-2\sqrt{6}}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+2\sqrt{6}}{5}\\x=\frac{3-2\sqrt{6}}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{3+2\sqrt{6}}{5};\frac{3-2\sqrt{6}}{5}\right\}\)
i) Ta có: \(6x^2-9x+40=0\)
\(\Leftrightarrow6\left(x^2-\frac{3}{2}x+\frac{20}{3}\right)=0\)
\(\Leftrightarrow x^2-\frac{3}{2}x+\frac{20}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{4}+\frac{9}{16}+\frac{293}{48}=0\)
\(\Leftrightarrow\left(x-\frac{3}{4}\right)^2+\frac{293}{48}=0\)(vô lý)
Vậy: \(S=\varnothing\)
DN
0
T
11 tháng 4 2019
a. Rút gọn đa thức và sắp xếp theo thứ tự giảm dần của biến..
\(A\left(x\right)=13x^4+3x^2+15x+7x^2-10x^4-7x-6-8x+15\)
\(=\left(13x^4-10x^4\right)+\left(3x^2+7x^2\right)+\left(15x-7x-8x\right)+\left(15-6\right)\)
\(=3x^4+10x^2+9.\)
\(B\left(x\right)=5x^4+10-5x^2-18+3x-10x^2-3x-4x^4\)
\(=\left(5x^4-4x^4\right)+\left(-5x^2-3x^2\right)+\left(3x-3x\right)+\left(10-18\right)\)
\(=x^4-8x^2-8\)
b. Tính M = A(x) + B(x) ; N = A(x) - B(x)
\(M=A\left(x\right)+B\left(x\right)=\left(3x^4+10x^2+9\right)+\left(x^4-8x^2-8\right)\)
\(=\left(3x^4+x^4\right)+\left(10x^2-8x^2\right)+\left(10-8\right)\)
\(=4x^4+2x^2+2\)
\(N=A\left(x\right)-B\left(x\right)=\left(3x^4+10x^2+9\right)-\left(x^4-8x^2-8\right)\)
\(=3x^4+10x^2+9-x^4+8x^2+8\)
\(=\left(3x^4-x^4\right)+\left(10x^2+8x^2\right)+\left(9+8\right)\)
\(=2x^4+18x^2+17\)
15 tháng 8 2019
Vì \(x=9\Rightarrow x+1=10\)
Thay x+1=10 vào biểu thức C ta dduojcw :
\(C=x^{14}-\left(x+1\right)x^{13}+\left(x+1\right)x^{12}-...-\left(x+1\right)x+10\)
\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-...-x^2-x+10\)
\(=-x+10\)
\(=-9+10\)
\(=1\)