Ta có :
/\(2x-1\)/\(\ge0\)
=> \(-\)/\(2x-1\)/ \(\le0\)(1)
=> \(-\)/\(2x-1\)/ \(-\frac{1}{2}\le-\frac{1}{2}\)
=> M \(\le-\frac{1}{2}\)(2)
Từ 1 và 2 ta có : Mmax = \(-\frac{1}{2}\)tại \(2x-1=0\).
=> \(x=\frac{1}{2}\)
Vậy Mmax khi \(x=\frac{1}{2}\)
Tìm GTLN
a) \(M=-\left|2x-1\right|-\frac{1}{2}\)
Co hội có L_I_K_E
Tìm x và y biết:
d)\(-1\frac{2}{3}-\left(\left|2x\right|+\frac{5}{6}\right)=\)\(-2\)e)\(\left(-\frac{1}{2}+\frac{1}{3}\right):\left|1-2x\right|-1\frac{1}{4}:\left(-\frac{5}{8}\right).\left(-\frac{1}{2}\right)^2=\frac{1}{3}\)
c)\(\left|2x-1\right|+\left|2y+1\right|+\left|2x-y\right|=0\)b)\(\left|2x-1\right|=2x-1\)
a)\(\left|x-3\right|=x+4\)
5. Tìm GTLN
a) M= 7-4|x+3|
b) N= \(\dfrac{18}{\left|x-2\right|+9}\)+5
Tìm giá trị lớn nhất của các biểu thức:
a,\(A=-x^2\)
b,\(B=-xx^2+5\)
c, \(C=6-x^4\)
d,\(D=\frac{1}{x^2+9}\)
e,\(E=\frac{1}{\left(x-1\right)^2+2}\)
f, \(F=11-\left(2x-3\right)^2\)
g,\(G=\frac{1}{\left|4-1\right|+3}\)
h,\(H=15-\left(2x-1\right)^8-\left|1-2y\right|\)
k,\(K=\frac{1-\left|8x-\frac{2}{3}\right|}{2}\)
Tìm x:
a)\(2.\left(3x-\frac{1}{2}\right)-2x=\frac{1}{2}\left(2x-3\right)\)
b)\(\left(2x-\frac{3}{5}\right)^2=\frac{4}{25}\)
c)\(\left(3x-1\right)^3=27\)
d)\(5-\left|x\right|=2\)
e)|2x+1|-3=3
f)|3-2x|=5
\(\left(5-x\right)\left(3x-\frac{1}{4}\right)=0\)
Tìm x ;
\(\left(\frac{1}{2}x^2\left(2x-1\right)^m-\frac{1}{2}x^{m+2}\right):\frac{1}{2}x^2=0\left(m\in N\right)\)
b)\(\left(1,78^{2x-2}-1,78^x\right):1,78^x=0\)
Tìm x, biết:
a)\(-3\frac{1}{2}+\frac{1}{3}.\left(x-1\right)=-1\frac{1}{3}:2\frac{1}{3}\)
b)\(\frac{x-1}{4}=\frac{x-2}{5}\)
c)\(\frac{x}{0,04}=\frac{0,09}{x}\)
d)\(4:\left(x-1\right)=\left(x-1\right):9\)
e)\(3\frac{1}{2}:x.\frac{1}{2}=5\frac{1}{3}:\frac{1}{2}.\frac{1}{5}\)
f)/2x-3/=5
g)\(\left(2x-\frac{1}{3}\right)^2=\left(1=3x\right)^2\)
h) \(\left(3-\frac{1}{2}:x\right)^2=\frac{1}{4}\)
i) \(\left(3-1\frac{1}{2}:x\right)^3=\frac{1}{8}\)
tìm x biết
a,\(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)
b,\(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)
c,\(\left|x-1\right|+4=6\)
d, \(\frac{x}{3}=\frac{y}{5}\) và y-x=24
e, \(\left(x^2-3\right)^2=16\)
f, \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\),
g, \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)
k,\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)
l,\(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
tìm giá trị của x để :
A=(13+x)(17+x)(2-x)\(\le\)0
B\(=\frac{\left(2x+6\right)\left(15-x\right)}{x+5}\ge0\)
C=\(\left(x^2+10\right)\left(2x-3\right)\left(1-4x\right)\le0\)
D=\(x^2\le5x\)
I x-12 I -x = 4
I 3x+1 I -2x = 3
5 - I x + 1 I =2
giúp mk nhé . ai nhanh nhất mk k cho 3k
