\(x^3-x^2-x+1=0\)
\(x^3-x^2-x+1=0\)
\(x^3-x^2-x+1=0\)
\(\Rightarrow x^2\left(x-1\right)-1\left(x-1\right)=0\)
\(\Rightarrow\left(x^2-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow x=1\)
\(x^3-x^2-x+1=0\)
\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)
\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+1\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Kết luận
tìm x:
\(\left(2x-3\right)^2-4x^2=0\)
Ta có :
\(\left(2x-3\right)^2-4x^2=0\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x\right)^2=0\)
\(\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\)
\(\Leftrightarrow-3\left(4x-3\right)=0\)
\(\Leftrightarrow4x-3=0\)
\(\Rightarrow x=\dfrac{3}{4}\)
Kết luận
Giải phương trình:\(\dfrac{x-29}{1970}+\dfrac{x-27}{1972}+\dfrac{x-25}{1974}+\dfrac{x-23}{1976}+\dfrac{x-21}{1978}+\dfrac{x-19}{1980}=\dfrac{x-1970}{29}+\dfrac{x-1972}{27}+\dfrac{x-1974}{25}+\dfrac{x-1976}{23}+\dfrac{x-1978}{21}+\dfrac{x-1980}{19}\)
Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}\)\(=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)
\(\Leftrightarrow\left(\frac{x-29}{1970}-1\right)+\left(\frac{x-27}{1972}-1\right)+\left(\frac{x-25}{1974}-1\right)+\left(\frac{x-23}{1976}-1\right)+\left(\frac{x-21}{1978}-1\right)+\left(\frac{x-19}{1980}-1\right)\)\(=\left(\frac{x-1970}{29}-1\right)+\left(\frac{x-1972}{27}-1\right)+\left(\frac{x-1974}{25}-1\right)+\left(\frac{x-1976}{23}-1\right)+\left(\frac{x-1978}{21}-1\right)+\left(\frac{x-1980}{19}-1\right)\)
\(\Leftrightarrow\frac{x-1999}{1970}+\frac{x-1999}{1972}+\frac{x-1999}{1974}+\frac{x-1999}{1976}+\frac{x-1999}{1978}+\frac{x-1999}{1980}\)\(=\frac{x-1999}{29}+\frac{x-1999}{27}+\frac{x-1999}{25}+\frac{x-1999}{24}+\frac{x-1999}{21}+\frac{x-1999}{19}\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}\right)\)\(=\left(x-1999\right)\left(\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}-\frac{1}{23}-\frac{1}{21}-\frac{1}{19}\right)=0\)\(\Leftrightarrow\) \(x-1999=0\) (Vì ...khác 0)
\(\Leftrightarrow x=1999\)(thỏa mãn)
Vậy \(x=1999\)
Cho a,b,c >0; a+b+c=1. Chứng minh \(\sqrt{2a+1}+\sqrt{2b+1}+\sqrt{2c+1}< 4\)
Giai: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)=4\)
(4x + 1)(12x - 1)(3x + 2)(x + 1) = 4
[(4x + 1)(3x + 2)][(12x - 1)(x + 1)] = 4
(12x2 + 8x + 3x + 2)(12x2 + 12x - x - 1) = 4
(12x2 + 11x + 2)(12x2 + 11x - 1) = 4
Đặt 12x2 + 11x - 1 = y, ta có:
(y + 3)y = 4
\(\Rightarrow\) y2 + 3y = 4
\(\Rightarrow\) y2 + 3y - 4 = 0
\(\Rightarrow\) y2 - y + 4y - 4 = 0
\(\Rightarrow\) (y2 - y) + (4y - 4) = 0
\(\Rightarrow\) y(y - 1) + 4(y - 1) = 0
\(\Rightarrow\) (y + 4)(y - 1) = 0
\(\Rightarrow\) (12x2 + 11x - 1 + 4)(12x2 + 11x - 1 - 1) = 0
\(\Rightarrow\) (12x2 + 11x + 3)(12x2 + 11x - 2) = 0
1 ô tô đi đường 80km trong 1 thời gian đã định, 3/4 đường đầu đi với vận tốc lớn hơn vận tốc dự định 10km/h. Quãng đường còn lại chạy chậm hơn vận tốc dự định 15km/h. Tính thời gian dự định?
Giải: \(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)
Cho \(a\ge1\), \(b\ge4\), \(c\ge9\). Tìm max \(K=\dfrac{bc\sqrt{a-1}+ca\sqrt{b-4}+ab\sqrt{c-9}}{abc}\)
Tìm các số a và b sao cho x3+ax+b chia cho x+1 thì dư 7, chia cho x-3 thì dư -5
Giúp mình với
Ta có: \(\left\{{}\begin{matrix}f\left(x\right)=x^3+ax+b=\left(x+1\right).Q\left(x\right)+7\\f\left(x\right)=x^3+ax+b=\left(x-3\right).g\left(x\right)-5\end{matrix}\right.\)
Ta có mệnh đề đúng với mọi x nên ta xét lần lượt \(x=-1;x=3\)
\(\left\{{}\begin{matrix}f\left(-1\right)=\left(-1\right)^3+a.\left(-1\right)+b=7\\f\left(3\right)=3^3+3a+b=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b-a-1=7\\3a+b+17=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-10\\b=-2\end{matrix}\right.\)
Phân tích đa thức thành nhân tử:
a,\(h\left(x\right)=x^4+3x^3+3x^2+3x+2\)
b,\(B=ab\left(a-b\right)\left(c+1\right)+bc\left(b-c\right)\left(a+1\right)+c\left(a-c\right)\left(b+1\right)\)
\(h\left(x\right)=x^4+3x^3+3x^2+3x+2\\ \Leftrightarrow h\left(x\right)=x^4+2x^3+x ^3+2x^2+x^2+2x+x+2\\ \Leftrightarrow h\left(x\right)=\left(x^4+2x^3\right)+\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(x+2\right)\\ \Leftrightarrow h\left(x\right)=x^3\left(x+2\right)+x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)\\ \Leftrightarrow h\left(x\right)=\left(x+2\right)\left(x^3+x^2+x+1\right)\\ \Leftrightarrow h\left(x\right)=\left(x+2\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\\ \Leftrightarrow h\left(x\right)=\left(x+2\right)\left(x+1\right)\left(x^2+1\right)\)