(x3+64).(\(\left|x\right|\)+14)=0
Những câu hỏi liên quan
let P(x) be a polynomial of degree 3 and x1, x2, x3 are the solutions of P(x)=0. let \(\frac{P\left(\frac{1}{3}\right)-P\left(\frac{-1}{3}\right)}{P\left(0\right)}=8,\frac{P\left(\frac{1}{4}\right)-P\left(\frac{-1}{4}\right)}{P\left(0\right)}=9\)and x1+x2+x3 = 35. find the value of \(\frac{x2+x3}{x1}+\frac{x1+x3}{x2}+\frac{x1+x2}{x3}\)
\(Cho:x1,x2,x3,....,x9\in Z\)thỏa mãn:
\(x=\left(1+x1\right)\left(1+x2\right)\left(1+x3\right)......\left(1+x9\right)=\left(1-x1\right)\left(1-x2\right)\left(1-x3\right)........\left(1-x9\right)\)
\(CMR:x.x1.x2.x3......x9=0\)
a) \(\left(\left(-0,5\right)^3\right)^n\)=\(\frac{1}{64}\). b) \(\left(x^n\right)^2\)=\(x^{14}\) ( x khác 0 và x khác 1 )
1. Tìm \(m\in\left[-10;10\right]\) để pt \(\left(x^2-2x+m\right)^2-2x^2+3x-m=0\) có 4 ng pb
2. Cho biết x1,x2 là nghiệm của pt \(x^2-x+a=0\) và x3,x4 là nghiệm của pt \(x^2-4x+b=0\) . Biết rằng \(\dfrac{x2}{x1}=\dfrac{x3}{x2}=\dfrac{x4}{x3}\), b >0 . Tìm a
1.
Đặt \(x^2-2x+m=t\), phương trình trở thành \(t^2-2t+m=x\)
Ta có hệ \(\left\{{}\begin{matrix}x^2-2x+m=t\\t^2-2t+m=x\end{matrix}\right.\)
\(\Rightarrow\left(x-t\right)\left(x+t-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=t\\x=1-t\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=x^2-2x+m\\x=1-x^2+2x-m\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m=-x^2+3x\\m=-x^2+x+1\end{matrix}\right.\)
Phương trình hoành độ giao điểm của \(y=-x^2+x+1\) và \(y=-x^2+3x\):
\(-x^2+x+1=-x^2+3x\)
\(\Leftrightarrow x=\dfrac{1}{2}\Rightarrow y=\dfrac{5}{4}\)
Đồ thị hàm số \(y=-x^2+3x\) và \(y=-x^2+x+1\):
Dựa vào đồ thị, yêu cầu bài toán thỏa mãn khi \(m< \dfrac{5}{4}\)
Mà \(m\in\left[-10;10\right]\Rightarrow m\in[-10;\dfrac{5}{4})\)
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a) \(\left(x^2+3x+2\right)^2=\left(x^2-x-2\right)^2\)
b) x3 + x2 - 4x - 4 = 0
a) Ta có: \(\left(x^2+3x+2\right)^2=\left(x^2-x-2\right)^2\)
\(\Leftrightarrow\left(x^2+3x+2\right)^2-\left(x^2-x-2\right)^2=0\)
\(\Leftrightarrow\left(x^2+3x+2-x^2+x+2\right)\left(x^2+3x+2+x^2-x-2\right)=0\)
\(\Leftrightarrow\left(4x+4\right)\left(2x^2+2x\right)=0\)
\(\Leftrightarrow4\left(x+1\right)\cdot2x\cdot\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: S={0;-1}
b) Ta có: \(x^3+x^2-4x-4=0\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
Vậy: S={-1;2;-2}
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Tìm x , biết
\(\left(\frac{1}{64}\right)^x\) = \(\left(-\frac{1}{8}\right)^{14}\)
\(\left(\frac{1}{64}\right)^x=\left(-\frac{1}{8}\right)^{14}\)
\(\left(\frac{1}{64}\right)^x=\left[\left(-\frac{1}{8}\right)^2\right]^7\)
\(\left(\frac{1}{64}\right)^x=\left(\frac{1}{64}\right)^7\)
\(\Rightarrow x=7\)
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26 tháng 7 2021 lúc 14:37
tìm m để pt \(x^4-2\left(m+1\right)x^2+2m+1=0\) có 4 nghiệm phân biệt
thỏa mãna, x1<x2<x3<X4<3
b,x1-x3=x3-x2=x2-x1
\(x^4-1-2\left(m+1\right)x^2+2\left(m+1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+1\right)-2\left(m+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-2m-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=2m+1\end{matrix}\right.\)
Pt có 4 nghiệm pb khi: \(\left\{{}\begin{matrix}2m+1>0\\2m+1\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{2}\\m\ne0\end{matrix}\right.\)
Do \(x=\pm1< 3\) nên để \(x_1< x_2< x_3< x_4< 3\) thì:
\(\sqrt{2m+1}< 3\Leftrightarrow m< 4\) \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{2}< m< 4\\m\ne0\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}x_1-x_3=x_3-x_2\\x_1-x_3=x_2-x_1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=-x_2\\x_1-x_3=-x_1-x_1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_2=-x_1\\x_3=3x_1\end{matrix}\right.\)
Do vai trò \(x_1;x_2\) như nhau, giả sử \(x_1< 0\) \(\Rightarrow x_1;x_3\) là 2 nghiệm âm
TH1: \(\left\{{}\begin{matrix}x_1=-1\\x_2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_3=-\sqrt{2m+1}\\x_3=3x_1\end{matrix}\right.\) \(\Rightarrow-\sqrt{2m+1}=-3\Rightarrow m=4\)
TH2: \(x_1=-\sqrt{2m+1}\Rightarrow\left\{{}\begin{matrix}x_3=-1\\x_3=3x_1\end{matrix}\right.\) \(\Rightarrow-1=-3\sqrt{2m+1}\) \(\Rightarrow m=-\dfrac{4}{9}\)
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Tìm x có giá trị nguyên để: (2x−3)(x−14)<0(2x−3)(x−14)<0\left(2x-3\right)\left(x-\dfrac{1}{4}\right)< 0
1,tìm xa.3^x81b.5^{x+2}125c.2^3*2^{x-1}64d. 7*7^{n+1}343e.left(x+3right)^5243f.left(2x-3right)^664g.left(x-6right)^3left(x-6right)^2h.left{left[left(2x+14right):2^2-3right]:3right}-10i.left{left[left(x:3+17right):10+3cdot2^4right]:10right}52,tínha.7^{3^1} b.7^{1^3} c.6^{1^{2^{3^4}}} d.2017^{2^{0^{1^0}}} e.6^3:6^2-6^2:6 f.frac{11cdot2^{22}cdot3^7-9^{15}}{left(2cdot3^{14}right)^2}
Đọc tiếp
1,tìm x
a.\(3^x\)=81
b.\(5^{x+2}\)=125
c.\(2^3\)*\(2^{x-1}\)=64
d. 7*\(7^{n+1}\)=343
e.\(\left(x+3\right)^5\)=243
f.\(\left(2x-3\right)^6\)=64
g.\(\left(x-6\right)^3\)=\(\left(x-6\right)^2\)
h.\(\left\{\left[\left(2x+14\right):2^2-3\right]:3\right\}\)-1=0
i.\(\left\{\left[\left(x:3+17\right):10+3\cdot2^4\right]:10\right\}=5\)
2,tính
a.\(7^{3^1}\) b.\(7^{1^3}\) c.\(6^{1^{2^{3^4}}}\) d.\(2017^{2^{0^{1^0}}}\) e.\(6^3:6^2-6^2:6\) f.\(\frac{11\cdot2^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)
1, Ta có :
a . 81 = 34 => 3x= 34 => x = 4 .
b. 125 = 53 => 5x+2 = 53 =>x + 2 = 3 => x = 1
c. 23 * 2x - 1 = 64
=> 23 + ( x - 1 ) = 64 = 26
=> 3 + ( x - 1 ) = 6
=> x - 1 = 6 - 3 = 3
x = 3 + 1
x = 4
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