x^2 - 27
giải các bất phương trình
a) \(27^{2-x}\le9\)
b) \(7^{3-x}< 49\)
c) \(27^{3-x}>9\)
d) \(2^{3-x}< 2^3\)
e) \(27^{3-x^2}< 27^{x+1}\)
a: \(27^{2-x}< =9\)
=>\(\left(3^3\right)^{2-x}< =3^2\)
=>\(3^{6-3x}< =3^2\)
=>6-3x<=2
=>-3x<=-4
=>\(x>=\dfrac{4}{3}\)
b: \(7^{3-x}< 49\)
=>\(7^{3-x}< 7^2\)
=>3-x<2
=>-x<2-3=-1
=>x>1
c: \(27^{3-x}>9\)
=>\(\left(3^3\right)^{3-x}>3^2\)
=>\(3^{9-3x}>3^2\)
=>9-3x>2
=>-3x>-7
=>\(x< \dfrac{7}{3}\)
d: \(2^{3-x}< 2^3\)
=>3-x<3
=>-x<0
=>x>0
e: \(27^{3-x^2}< 27^{x+1}\)
=>\(3-x^2< x+1\)
=>\(-x^2-x+2< 0\)
=>\(x^2+x-2>0\)
=>(x+2)(x-1)>0
=>\(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
Giaỉ các phương trình sau
\(a,\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)\(a,\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
nên \(x^2-10x-2000=0\)
\(\Leftrightarrow x^2+40x-50x-2000=0\)
\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Vậy: S={-40;50}
cho minh hoi vi sao lai đổi 27/4 thanh 27/2
2[x^2+2x5/2+(5/2)^2-27/4)
2(x+5/2)^2-27/2
vì còn 2 nhân với 27/4 sau khi phá ngoặc : 2.27/4 = 27/2
vậy nên 27/4 thành 27/2
Tính giá trị biểu thức:27 x 2 + 5 x 27 + 27 x 3
27 x 2 + 5 x 27 + 27 x 3
= 27 x (2 + 5 + 3)
= 27 x 10
= 270
a) 2019/2020 x 4/11 + 2019/2020 x 5/11 + 2019/2020 x 2/11
b) 17/14 x 25/27 - 1/14 x 25/27 - 2/14 x 25/27
\(a.=\dfrac{2019}{2020}\times\left(\dfrac{4}{11}+\dfrac{5}{11}+\dfrac{2}{11}\right)\\ =\dfrac{2019}{2020}\times1=\dfrac{2019}{2020}\\ b.=\dfrac{25}{27}\times\left(\dfrac{17}{14}-\dfrac{1}{14}-\dfrac{2}{14}\right)\\ =\dfrac{25}{27}\times1=\dfrac{25}{27}\)
Tìm x, biết:
a) x × 3 = 27
4 × x = 20
10 + x : 2 = 20
b) x × 3 = 27 + 3
27 : x = 789 - 780
a. x × 3 = 27
x = 27 : 3
x = 9
b. 4 × x = 20
x = 20 : 4
x = 5
c. 10 + x : 2 = 20
x : 2 = 20 – 10
x : 2 = 10
x = 10 × 2
x = 20
d. x × 3 = 27 + 3
x × 3 = 30
x = 30 : 3
x = 10
e. 27 : x = 789 – 780
27 : x = 9
x = 27 : 9
x = 3
a, x^3-4x^2-12x+27
b, (x^2+x+1) * (x^2+x+2)-27
tìm x:
2(x-5)+3=-7
4x+(27+2^3)=1
/x-3/+5=-27
-3/x-1/=-27
2(x-5)+3=-7
=>2x-10=-7+3=-4
=>2x=-4+10=6
=>x=6:2=3
4x+(27+2^3)=1
=>4x+(27+8)=1
=>4x+35=1
=>4x=1-35=-34
=>x=-34:4=8.5
lx-3l+5=-27
=>lx-3l=-27-5=-32
mà lx-3l>0 với mọi x
=>k co gt x t/m
-3lx-1l=-27
=>lx-1l=-27:(-3)=9
=>x-1=+9
+)x-1=9=>x=10
+)x-1=-9=>x=-8
\(\dfrac{\sqrt{27+x^2+x}}{2+\sqrt{5-\left(x^2+x\right)}}=\dfrac{\sqrt{27+2x}}{2+\sqrt{5-2x}}\)
x^3+27=27=9-x^2
x^3+27=27
x=0
=>27=9-0^2
=>27=9
=>x không tồn tại