giai phuong trinh
x+2/2016+x+3/2015+x+4/2014+x+2036/6=0
Những câu hỏi liên quan
(2x^+x-2015)^2+4(2x^+x-2016)^2=4(2x^+x-2015)(2x^+x-2016) giai phuong trinh
Sửa đề:
\((2x^2+x-2015)^2+4(x^2-5x-2016)^2=4(2x^2+x-2015)(x^2-5x-2016)\)
\(\Rightarrow\left(2x^2+x-2015\right)^2-2.\left(2x^2+x-2015\right).2.\left(x^2-5x-2016\right)+[2.\left(x^2-5x-2016\right)]^2=0\)
\(\Rightarrow[2x^2+x-2015-2.\left(x^2-5x-2016\right)]^2=0\)
\(\Rightarrow11x+2017=0\)
\(\Rightarrow x=\frac{-2017}{11}\)
Giai cac Phuong trinh sau: x+1 / 2014 +x+2 / 2013=x+3 / 2012 +x+4 / 2011
Giai phuong trinh
a) (x+1)^4+(x-3)^4=0
b) x^4 + 2x^3 - 4x^2 -5x -6=0
a) Ta có: \(\left(x+1\right)^4+\left(x-3\right)^4=0\)
Nhận thấy: \(\hept{\begin{cases}\left(x+1\right)^4\ge0\left(\forall x\right)\\\left(x-3\right)^4\ge0\left(\forall x\right)\end{cases}\Rightarrow}\left(x+1\right)^4+\left(x-3\right)^4\ge0\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+1\right)^4=0\\\left(x-3\right)^4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\x=3\end{cases}}\) (mâu thuẫn)
=> pt vô nghiệm
b) \(x^4+2x^3-4x^2-5x-6=0\)
\(\Leftrightarrow\left(x^4-2x^3\right)+\left(4x^3-8x^2\right)+\left(4x^2-8x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+4x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x^3+3x^2\right)+\left(x^2+3x\right)+\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)
Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\forall x\right)\)
=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
a,\(\left(x+1\right)^4+\left(x-3\right)^4=0\)
\(x^4-1+x^4-81=0\)
\(2x^4-82=0\)
\(2x^4=82\)
\(x^4=41\)
\(x=\sqrt[4]{41}\)
\(\Rightarrow\)vô nghiệm
1 . tong binh phuong tat ca cac nghiem cua phuong trinh : x4(x-1)+(x-1)x3=0
2.neu da thuc 2m3-m+n co 2 nhan tu la : x+2 va x-1 thi gia tri cua 2m+3n la
3.tong x; y thoa man (x-2015)2+(y-2014)4>hoac bang 0 la ?
tinh nhanh bieu thuc sau
a) x^6-20x^5-20x^4-20x^4-20x^3-20x^2-20x+1992 voi x= 21
b)x^7-26x^6+27x^5-47x^4-77x^3+50x^2+x+1989 tai x = 25
c)4029-2015x+2015x^2-....+2015x^2014-x^2015 tai x= 2014
d)A= (x-1/x-2015+x-2/x-2014+x-3/x-2013+....+x-2014/x-2+x-2015/x-1):(1/2 +1/3+1/4+....+1/2016) tai x=2016 dau / o day la phan do nha!!
xin nho cac ban giai giup cho minh! cam on nhieu!!!!!!!! minh dang can ket qua gap
Giai phuong trinh:
\(x^4+\sqrt{x^2+2015}=2015\)
=> \(x^4-2015+\sqrt{x^2+2015}=0\)
<=> \(x^4-\left(x^2+2015\right)+x^2+\sqrt{x^2+2015}=0\)
<=> \(\left(x^2+\sqrt{x^2+2015}\right).\left(x^2-\sqrt{x^2+2015}\right)+\left(x^2+\sqrt{x^2+2015}\right)=0\)
<=> \(\left(x^2+\sqrt{x^2+2015}\right).\left(x^2-\sqrt{x^2+2015}+1\right)=0\)
=> \(x^2-\sqrt{x^2+2015}+1=0\) (*) (Vì \(x^2+\sqrt{x^2+2015}>0\) với mọi x )
Đặt \(\sqrt{x^2+2015}=t\Rightarrow x^2+2015=t^2\Rightarrow x^2=t^2-2015\)
thay vào (*) ta được: t2 - 2015 - t + 1 = 0
=> t2 - t - 2014 = 0
\(\Delta\) = 1 + 4. 2014 = 8057
=> \(t_1=\frac{1+\sqrt{8057}}{2};t_2=\frac{1-\sqrt{8057}}{2}\)
nhận t1 => x2 = \(\left(\frac{1+\sqrt{8057}}{2}\right)^2-2015\) => x = .....
Đúng 1
Bình luận (0)
Giai phuong trinh
-x^3 + x^2 +4 =0
-x3 + x2 + 4 = 0
<=> -(x - 2)(x2 + x + 2) = 0
<=> x - 2 = 0
x = 0 + 2
x = 2
Mà vì x2 + x + 2 # 0
=> x = 2
Đúng 0
Bình luận (0)
giai cac phuong trinh sau
a) x + 5 can x - 6 =0
b) x - can x + 1/4 = 0
a: \(x+5\sqrt{x}-6=0\)
\(\Leftrightarrow\sqrt{x}-1=0\)
hay x=1
b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)
hay \(x=\dfrac{1}{4}\)
Đúng 0
Bình luận (1)
giai cac phuong trinh sau
a) x + 5 can x - 6 =0
b) x - can x + 1/4 = 0
a: \(x+5\sqrt{x}-6=0\)
\(\Leftrightarrow\sqrt{x}-1=0\)
hay x=1
b: \(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\Leftrightarrow\sqrt{x}-\dfrac{1}{2}=0\)
hay \(x=\dfrac{1}{4}\)
Đúng 2
Bình luận (1)