b. + Vì \(|6-2x|\ge0\)\(\forall x\)

\(\Rightarrow\)\(|6-2x|-5\ge0-5\)\(\forall x\)

\(\Rightarrow\)B\(\ge\)-5 \(\forall x\)

Vậy GTNN của B= -5 \(\Leftrightarrow\)6-2x=0

                                    \(\Leftrightarrow\)2x=6

                                   \(\Leftrightarrow\)x=3

+ Vì -\(|6-2x|\le0\forall x\)

\(\Rightarrow\)\(|6-2x|-5\le0+5\forall x\)

\(\Rightarrow B\le5\forall x\)

Vậy GTLN của B= 5 \(\Leftrightarrow6-2x=0\)

                                \(\Leftrightarrow2x=1\)

                                \(\Leftrightarrow x=\frac{1}{2}\)

c,+ Vì \(|x+1|\ge0\forall x\)

\(\Rightarrow\)\(3-|x+1|\ge3-0\forall x\)

\(\Rightarrow C\ge3\forall x\)

Vậy GTNN của C=3 \(\Leftrightarrow x+1=0\)

                                 \(\Leftrightarrow x=-1\)

+ Vì \(-|x+1|\le0\forall x\)

\(\Rightarrow3-|x+1|\le3+0\forall x\)

\(\Rightarrow C\le3\forall x\)

Vậy GTLN của \(C=3\Leftrightarrow x+1=0\)

                                     \(\Leftrightarrow x=-1\)

Mình chỉ làm vậy thôi nhé!

THANKS  BẠN NHIỀU NHA

không có j

Tự học giúp bạn có được một gia tài
Jim Rohn – Triết lý cuộc đời

a) 5.2² + (x + 3) = 5²

5.4 + x + 3 = 25

20 + x + 3 = 25

x + 23 = 25

x = 25 - 23

x = 2

b) 2³ + (x - 3²) = 5³ - 4³

8 + (x - 9) = 125 - 64

8 + x - 9 = 61

x - 1 = 61

x = 61 + 1

x = 62

c) 4.(x - 5) - 2³ = 2⁴.3

4x - 20 - 8 = 16.3

4x - 28 = 48

4x = 48 + 28

4x = 76

x = 76 : 4

x = 19

d) 5.(x + 7) - 10 = 2³.5

5x + 35 - 10 = 8.5

5x + 25 = 40

5x = 40 - 25

5x = 15

x = 15 : 5

x = 3

e) 7² - 7.(13 - x) = 14

49 - 91 + 7x = 14

7x - 42 = 14

7x = 14 + 42

7x = 56

x = 56 : 7

x = 8

a) \(5\cdot2^2+\left(x+3\right)=5^2\)

\(\Rightarrow x+3=5^2-5\cdot2^2\)

\(\Rightarrow x+3=25-5\cdot4\)

\(\Rightarrow x+3=5\)

\(\Rightarrow x=5-3\)

\(\Rightarrow x=2\)

b) \(2^3+\left(x-3^2\right)=5^3-4^3\)

\(\Rightarrow8+\left(x-9\right)=125-64\)

\(\Rightarrow8+x-9=61\)

\(\Rightarrow x-1=61\)

\(\Rightarrow x=61+1\)

\(\Rightarrow x=62\)

c) \(4\left(x-5\right)-2^3=2^4\cdot3\)

\(\Rightarrow4\left(x-5\right)=2^4\cdot3+2^3\)

\(\Rightarrow4\cdot\left(x-5\right)=16\cdot3+8\)

\(\Rightarrow4\cdot\left(x-5\right)=56\)

\(\Rightarrow x-5=56:4\)

\(\Rightarrow x-5=14\)

\(\Rightarrow x=19\)

d) \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Rightarrow5\left(x+7\right)=8\cdot5+10\)

\(\Rightarrow5\left(x+7\right)=40+10\)

\(\Rightarrow5\left(x+7\right)=50\)

\(\Rightarrow x+7=10\)

\(\Rightarrow x=10-7\)

\(\Rightarrow x=3\)

e) \(7^2-7\left(13-x\right)=14\)

\(\Rightarrow7\left(13-x\right)=7^2-14\)

\(\Rightarrow7\left(13-x\right)=49-14\)

\(\Rightarrow7\left(13-x\right)=35\)

\(\Rightarrow13-x=5\)

\(\Rightarrow x=13-5\)

\(\Rightarrow x=8\)

f) \(5x-5^2=10\)

\(\Rightarrow5x=10+5^2\)

\(\Rightarrow5x=10+25\)

\(\Rightarrow5x=35\)

\(\Rightarrow x=\dfrac{35}{5}\)

\(\Rightarrow x=7\)

g) \(9x-2\cdot3^2=3^4\)

\(\Rightarrow9x=3^4+2\cdot3^2\)

\(\Rightarrow9x=81+2\cdot9\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=\dfrac{99}{9}\)

\(\Rightarrow x=11\)

h) \(10x+2^2\cdot5=10^2\)

\(\Rightarrow10x=10^2-2^2\cdot5\)

\(\Rightarrow10x=100-4\cdot5\)

\(\Rightarrow10x=80\)

\(\Rightarrow x=\dfrac{80}{10}\)

\(\Rightarrow x=8\)

i) \(125-5\left(4+x\right)=15\)

\(\Rightarrow5\left(4+x\right)=125-5\)

\(\Rightarrow5\left(4+x\right)=120\)

\(\Rightarrow4+x=\dfrac{120}{5}\)

\(\Rightarrow4+x=24\)

\(\Rightarrow x=24-4\)

\(\Rightarrow x=20\)

j) \(2^6+\left(5+x\right)=3^4\)

\(\Rightarrow5+x=3^4-2^6\)

\(\Rightarrow5+x=81-64\)

\(\Rightarrow5+x=17\)

\(\Rightarrow x=17-5\)

\(\Rightarrow x=12\)

f) 5x - 5² = 10

5x - 25 = 10

5x = 10 + 25

5x = 35

x = 35 : 5

x = 7

g) 9x - 2.3² = 3⁴

9x - 2.9 = 81

9x - 18 = 81

9x = 81 + 18

9x = 99

x = 99 : 9

x = 11

h) 10x - 2².5 = 10²

10x - 4.5 = 100

10x - 20 = 100

10x = 100 + 20

10x = 120

x = 120 : 10

x = 12

i) 125 - 5.(4 + x) = 15

5.(4 + x) = 125 - 15

5.(4 + x) = 110

4 + x = 110 : 5

4 + x = 22

X = 22 - 4

x = 18

j) 2⁶ + (5 + x) = 3⁴

64 + 5 + x = 81

69 + x = 81

x = 81 - 69

x = 12

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

4,  Q = |x+\(\frac{1}{5}\) | -x +\(\frac{4}{7}\)

 xét x \(\ge\) \(-\frac{1}{5}\)

 Ta Có  Q = |x+\(\frac{1}{5}\) | -x + \(\frac{4}{7}\)  = x+\(\frac{1}{5}\) - x +\(\frac{4}{7}\) = \(\frac{27}{35}\)   (1)

xét x \(< -\frac{1}{5}\)

Ta có Q = | x +\(\frac{1}{5}\) | - x + \(\frac{4}{7}\) = -x - \(\frac{1}{5}\) - x + \(\frac{4}{7}\) = -2x  + \(\frac{13}{35}\)

với x \(< -\frac{1}{5}\) 

=> -2x \(>\) \(\frac{2}{5}\) 

=> -2x + \(\frac{13}{35}\) \(>\frac{27}{35}\) (2)

Từ (1) và (2) => MinQ = \(\frac{27}{35}\) khi \(x\ge-\frac{1}{5}\)

5 ,  D = |x| + |8-x| 

D = |x| + |8-x| \(\ge\) |x+8-x|  = |8| = 8

Dấu ''='' xảy ra khi   x(8-x) \(\ge\) 0  <=> 0\(\le\)x\(\le\) 8 

Vậy MinD = 8 khi \(0\le x\le8\) 

6,L=  |x - 2012| + |2011 - x| 

L = |x-2012| + |2011-x| \(\ge\) | x-2012 + 2011 - x |  = |-1| = 1 

Dấu ''= '' xảy ra khi ( x-2012)(2011-x) \(\ge\) 0  

làm nốt câu 6 nãy ấn nhầm 

<=> 2011\(\le\) x \(\le\) 2012

Vậy MinL = 1 khi \(2011\le x\le2012\) 

7 , E = | x- \(\frac{2006}{2007}\) | + |x-1| 

Ta có :

E = |x-\(\frac{2006}{2007}\) | + |1-x| 

E = | x - \(\frac{2006}{2007}\) | + |1-x| \(\ge\) | x - \(\frac{2006}{2007}\) + 1 - x |  = \(\frac{1}{2007}\) 

Dấu ''='' xảy ra khi (x- \(\frac{2006}{2007}\) ) ( 1-x ) \(\ge0\) <=>  \(\frac{2006}{2007}\le x\le1\) 

Vậy MinE = \(\frac{1}{2007}\) khi \(\frac{2006}{2007}\le x\le1\) 

8 ,F = | x -\(\frac{1}{4}\) | + | \(x-\frac{3}{4}\) | 

Ta có :

F  = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\)   - x | 

F  = | x - \(\frac{1}{4}\) | + | \(\frac{3}{4}\) -x | \(\ge\) | x - \(\frac{1}{4}\) + \(\frac{3}{4}\) -x  |  = \(\frac{1}{2}\) 

Dấu ''='' xảy ra khi ( x-\(\frac{1}{4}\) ) ( \(\frac{3}{4}-x\) ) \(\ge\) 0    <=>  \(\frac{1}{4}\le x\le\frac{3}{4}\) 

Vậy MinF = \(\frac{1}{2}\) khi \(\frac{1}{4}\le x\le\frac{3}{4}\)

\(C=-\left|x+\frac{4}{7}\right|+\frac{12}{19}\)

Ta có: \(\left|x+\frac{4}{7}\right|\ge0\)nên \(-\left|x+\frac{4}{7}\right|\le0\)

\(\Rightarrow C=-\left|x+\frac{4}{7}\right|+\frac{12}{19}\le\frac{12}{19}\)

\(\Rightarrow C_{max}=\frac{12}{19}\)

(Dấu "="\(\Leftrightarrow x=\frac{-4}{7}\))

\(D=\left|x-\frac{5}{7}\right|+\frac{2}{3}\)

Vì \(\left|x-\frac{5}{7}\right|\ge0\)nên \(D=\left|x-\frac{5}{7}\right|+\frac{2}{3}\ge\frac{2}{3}\)

\(\Rightarrow D_{min}=\frac{2}{3}\)

(Dấu "="\(\Leftrightarrow x=\frac{5}{7}\))

\(F=\left|x-20\right|+\left|x-2015\right|\)

\(\Rightarrow F=\left|x-20\right|+\left|2015-x\right|\ge\left|\left(x-20\right)+\left(2015-x\right)\right|\)

\(=\left|2015-20\right|=\left|1995\right|=1995\)

\(\Rightarrow F_{min}=1995\)

(Dấu "="\(\Leftrightarrow x\le2015\))

cái gì thế này???????????????????????????????????

mik lp 6 nhưng nhìn bài của bn mik ko hiểu j cả luôn ý