a) x:2,1=13,04-8,63

x:2,1=4,41

x=4,41x2,1=9,261

b)x:2,04=7,5x2,18

x:2,04=16,35

x=16,35x2,04=33,354

Bài 2 : 

a) \(A=3,7+\left|4,3-x\right|\ge3,7\)

Min A = 3,7 \(\Leftrightarrow x=4,3\)

b) \(B=\left|3x+8,4\right|-14\ge-14\)

Min B = -14 \(\Leftrightarrow x=\frac{-14}{5}\)

c) \(C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

Min C = 17,5 \(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{-3}{2}\end{cases}}\)

d) \(D=\left|x-2018\right|+\left|x-2017\right|\)

\(D=\left|2018-x\right|+\left|x-2017\right|\ge\left|2018-x+x-2017\right|=1\)

Min D =1 \(\Leftrightarrow\left(2018-x\right)\left(x-2017\right)\ge0\)

\(\Leftrightarrow2017\le x\le2018\)

\(A=3,7+\left|4,3-x\right|\)

Ta có \(\left|4,3-x\right|\ge0\Leftrightarrow A=3,7+\left|4,3-x\right|\ge3,7\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left|4,3-x\right|=0\Leftrightarrow4,3-x=0\Leftrightarrow x=4,3\)

\(B=\left|3x+8,4\right|-14\)

Ta có \(\left|3x+8,4\right|\ge0\Leftrightarrow B=\left|3x+8,4\right|-14\ge-14\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left|3x+8,4\right|=0\Leftrightarrow3x=-8,4\Leftrightarrow x=2,8\)

\(C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)

Ta có \(\hept{\begin{cases}\left|4x-3\right|\ge0\\\left|5y+7,5\right|\ge0\end{cases}}\Leftrightarrow C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

Dấu '' = '' xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|4x-3\right|=0\\\left|5y+7,5\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}\)

\(D=\left|x-2018\right|+\left|x-2017\right|\)

\(\Leftrightarrow D=\left|x-2018\right|+\left|2017-x\right|\)

Áp dụng bất đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)ta có

\(D\ge\left|x-2018+2017-x\right|=\left|-1\right|=1\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left(2017-x\right)\left(x-2018\right)\ge0\Leftrightarrow2018\ge x\ge2017\)

a: \(P\left(1\right)=1^3-1^2-4\cdot1+4=-4+4=0\)

=>x=1 là nghiệm của P(x)

\(P\left(-2\right)=\left(-2\right)^3-\left(-2\right)^2-4\cdot\left(-2\right)+4=-8-4+8+4=0\)

=>x=-2 là nghiệm của P(x)

b: \(P\left(1\right)=5\cdot1^3-7\cdot1^2+4\cdot1-2=5-7+4-2=0\)

=>x=1 là nghiệm của P(x)

\(a,A=\left|3,4-x\right|+1,7\ge1,7\)

Dấu \("="\Leftrightarrow3,4-x=0\Leftrightarrow x=3,4\)

\(c,C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}4x-3=0\\5y+7,5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-\dfrac{3}{2}\end{matrix}\right.\)

\(\dfrac{7,5+1,5+\dfrac{5}{4}+\dfrac{5}{11}}{10,5+2,1+\dfrac{7}{4}+\dfrac{7}{11}}+x=\dfrac{6}{7}\)

\(\dfrac{9+\dfrac{5}{4}+\dfrac{5}{11}}{\dfrac{63}{5}+\dfrac{7}{4}+\dfrac{7}{11}}+x=\dfrac{6}{7}\)

\(\dfrac{\dfrac{36}{4}+\dfrac{5}{4}+\dfrac{5}{11}}{\dfrac{252}{20}+\dfrac{35}{20}+\dfrac{7}{11}}+x=\dfrac{6}{7}\)

\(\dfrac{\dfrac{41}{4}+\dfrac{5}{11}}{\dfrac{287}{20}+\dfrac{7}{11}}+x=\dfrac{6}{7}\)

\(\dfrac{\dfrac{451}{44}+\dfrac{20}{44}}{\dfrac{3157}{220}+\dfrac{140}{220}}+x=\dfrac{6}{7}\)

\(\dfrac{\dfrac{471}{44}}{\dfrac{3297}{220}}+x=\dfrac{6}{7}\)

\(\dfrac{5}{7}+x=\dfrac{6}{7}\)

\(x=\dfrac{6}{7}-\dfrac{5}{7}\)

\(x=\dfrac{1}{7}\)

a) /4x - 3/ + /5y+7,5/ >= 0

=> C>= 17,5

=> C min = 17,5 <=> 4x-3 = 0 và 5y + 7,5 =0 <=> x = 3/4 và y = -3/2

b) Áp dụng /A/ = /-A/

=> D = /x-2001/ + /2002-x/

Lại áp dụng /a/ + /b/ >= /a+b/

=> D>= /x-2001+2002-x/ = 1

=> D min = 1 <=> (x - 2001)(2002 - x) >= 0 <=> 2001 <= x <= 2002

\(A=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)

Ta thấy \(\left|4x-3\right|\ge0;\left|5y+7,5\right|\ge0\)

\(\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

\(\Rightarrow A\ge17,5\)

Dấu "=" xảy ra  \(\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}\)

...
\(B=\left|x-2\right|+\left|x-6\right|+2017\)

\(=\left|x-2\right|+\left|6-x\right|+2017\)

Ta thấy \(\left|x-2\right|+\left|6-x\right|\ge\left|x-2+6-x\right|=4\)

\(\Rightarrow B\ge4+2017=2021\)

Dấu "=" xảy ra khi \(2\le x\le6\)

....

\(C=\left(2x+1\right)^{2020}-2019\)

Ta thấy \(\left(2x+1\right)^{2020}\ge0\)

\(\Rightarrow C=\left(2x+1\right)^{2020}-2019\ge-2019\)

Dấu "=" xảy ra khi \(2x+1=0\Leftrightarrow x=-\frac{1}{2}\)

....

a﴿Ta có: |4,3‐x|\(\ge\)0﴾với mọi x﴿
nên 3,7+|4,3‐x|\(\ge\)3,7 hay A\(\ge\)3,7
Do đó, GTNN của A là 3,7 khi:|4,3‐x|=0
4,3‐x=0
x=4,3‐0
x=4,3
b﴿Ta có: |2x‐1,5|>=0﴾với mọi x﴿
‐|2x‐1,5|<=0
nên 5,5‐|2x‐1,5|<=5,5 hay B<=5,5
Do đó, GTLN của B là 5,5 khi:|2x‐1,5|=0
2x‐1,5=0
2x=0+1,5
2x=1,5
x=1,5/2=15/2=7,5
Vậy GTLN của B là 5,5 khi x=7,5

c)ta có 4x − 3 ≥ 0; 5x + 7,5 ≥ 0 

⇒E ≥ 17,5

=>GTNN của C là 17,5 hi x₁=3/4 hoặc x₂=-1,5