theo mình thì đề thế này
Tìm số nguyên x để biểu thức sau đạt GTNN
\(A=\left(x-1\right)^2+2008\)
Giải
Vì \(\left(x-1\right)^2\ge0\\ \Rightarrow\left(x-1\right)^2+2008\ge2008\)
Dấu "=" xảy ra
\(\\ \left(x-1\right)^2=0\\ \Rightarrow x-1=0\\ \Rightarrow x=1\)
Ta có \(KL:A_{min}=2008\Rightarrow x=1\)
Tìm GTNN hoặc GTLN:
a) \(A=3,5+\left|X-2017\right|-9\)
b) \(B=\left|-2,5-3\right|-\left|x+1,5\right|\)
c) \(C=\left|x+1,5\right|-7,5+\left|y-3,2\right|\)
d) \(D=-\left|2x-x\right|-\left|4-y\right|+9,5\)
e) \(E=\left|x+2\right|+\left|x-y+1\right|^{2018}-2,7\)
f) \(E=\left|x+2\right|-\left|x-y+1\right|^{2018}-2,7\)
Tính GTLN hoặc GTNN và phân biệt đâu là GTLN và GTLN:
a) \(A=4-\left(x+1\right)^{2018}\)
b) \(B=\left(x-3\right)^2-2017\)
c) \(C=\dfrac{4}{\left(x+1\right)^2+2}\)
d) \(D=\left(2x-y+1\right)^{2018}+\left|y+1\right|+2017\)
+Tim x :
a ) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
b )\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
c ) \(\left(x+1\right)^{x+2}=\left(x-1\right)^{x+6}\) (x ϵ Z )
d ) \(\left(2x+3\right)^{2016}=\left(2x+3\right)^{2018}\)
e ) \(\frac{3}{x+2.x+5}+\frac{5}{x+5.x+10}+\frac{7}{x+10.x+17}=\frac{x}{x+2.x+17}\) Với \(x\notin\left\{-2;-5;-10;-17\right\}\)
\(\left(x-3\right)\cdot\left(y-2\right)=7\)
\(\left(x-1\right)\cdot\left(y-1\right)=2\)
\(\left(x-1\right)\cdot\left(y-2\right)=2\)
Tìm x:
e/ \(\left(x-5\right)^4=\left(x-5\right)^6\)
i/ \(\left(x+2\right)^5=2^{10}\)
k/ \(\left(x+1\right)^2=\left(x+1\right)^0\)
l/ \(\left(3x-4\right).\left(x-1\right)^3+0\)
Tìm x, y ϵ Z biết :
a) \(\left|x-1\right|+\left|y-2\right|=1\)
b) \(\left|x+3\right|+\left|y\right|=0\)
c) \(\left|1-x\right|+\left|y+2\right|=2\)
Câu 1 : Tìm x :
a) \(\left(-11\right)^2-15.\left(x-2\right)=134-16x\)
b) \(\left(4x+1\right)\left(x^2-16\right)=0\)
c) \(-2\left(x-3\right)+\left(-2\right)^2=4-3x\)
\(\frac{\left[\left(e-m\right)^2-\left(e+m^2\right)\right].\left[\left(y-1\right)^2-\left(y+1\right)^2\right]}{a.16.n.h}x\frac{ê}{u^{-1}}\)
bài 1 : Tìm x :
1. \(\left(81-x^2\right).\left(-2x-16\right).\left(-3x+15\right)=0\)
bài 2: Tìm x,y thuộc Z
1, \(\left(2x-1\right).\left(4y+2\right)=-42\)
2,\(\left(5x+1\right).\left(y-1\right)=4\)
