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1.
a, (x-5)2
Ta có x2 luôn \(\ge\) 0 với mọi x, suy ra: (x-5)2 \(\ge\) 0 với mọi x
Nên: (x-5)2 \(\ge\) 0 với mọi x
Suy ra: đa thức này không có nghiệm.
Bài easy quá mà!
4. a) Áp dụng tỉ dãy số bằng nhau:
\(\frac{a_1-1}{100}=\frac{a_2-2}{99}=...=\frac{a_{100}-100}{1}\)
\(=\frac{\left(a_1+a_2+...+a_{100}\right)-\left(1+2+...+100\right)}{100+99+...+2+1}=\frac{5050}{5050}=1\)
Suy ra: \(a_1-1=100\Leftrightarrow a_1=101\)
\(a_2-2=99\Leftrightarrow a_2=101\)
.......v.v...
\(a_{100}-100=1\Leftrightarrow a_{100}=101\)
Do đó: \(a_1=a_2=a_3=...=a_{100}=101\)
Bài 5/
Theo t/c dãy tỉ số bằng nhau,ta có: \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)\(=\frac{2x}{x}\)
Suy ra:
\(\frac{y+z-x}{x}=\frac{2x}{x}\Leftrightarrow y+z-x=2x\Rightarrow x=y=z\) (vì nếu \(x\ne y\ne z\Rightarrow y+z-x\ne2x\) "không thỏa mãn")
Thay vào A,ta có: \(A=\left(1+\frac{x}{x}\right)\left(1+\frac{y}{y}\right)\left(1+\frac{z}{z}\right)=2.2.2=8\)
a) \(x+2x+3x+...+100x=-213\)
\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)
\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)
\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)
d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)
c) \(3\left(x-2\right)+2\left(x-1\right)=10\)
\(\Rightarrow3x-6+2x-2=10\)
\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)
a) \(x+2x+3x+4x+...+100x=-213\)
\(x.\left(1+2+3+4+...+100\right)=-213\)
\(x.5050=-213\)
\(x=-\frac{213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)
\(\frac{1}{2}x=-\frac{43}{12}\)
\(x=\frac{-43}{6}\)
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
Bài 1:
\(M\left(1\right)=a+b+6\)
Mà \(M\left(1\right)=0\)
\(\Rightarrow a+b+6=0\)
\(\Rightarrow a+b=-6\)( * )
\(\Rightarrow2a+2b=-12\) (1)
Ta có: \(M\left(-2\right)=4a-2b+6\)
Mà \(M\left(-2\right)=0\)
\(\Rightarrow4a-2b=-6\)(2)
Lấy (1) cộng (2) ta được:
\(6a=-18\)
\(a=-3\)
Thay a=-3 vào (* ) ta được:
\(b=-3\)
Vậy a=-3 ; b=-3
Bài 2:
a) \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{8}-\frac{y}{4}=\frac{5}{x}\)
\(\Leftrightarrow\frac{1}{8}-\frac{2y}{8}=\frac{5}{x}\)
\(\Leftrightarrow\frac{1-2y}{8}=\frac{5}{x}\)
\(\Leftrightarrow\left(1-2y\right).x=5.8\)
\(\Leftrightarrow\left(1-2y\right).x=40\)
Vì \(x,y\in Z\Rightarrow1-2y\in Z\)
mà \(40=1.40=40.1=5.8=8.5=\left(-1\right).\left(-40\right)=\left(-40\right).\left(-1\right)=\left(-5\right).\left(-8\right)=\left(-8\right).\left(-5\right)\)
Thử từng TH