\(A=\left|x-3,5\right|+\left|4,1-x\right|\)

Ta có: \(3,5\le x\le4,1\)

\(\Rightarrow x-3,5+4,1-x=0,6\)

\(B=\left|-x+3,5\right|+\left|x-4,1\right|=\left|3,5-x\right|+\left|x-4,1\right|\)

Ta có: \(3,5\le x\le4,1\)

\(\Rightarrow x-3,5+4,1-x=0,6\)

`A= (x+5)/(x+3 )` 

Điều kiện: `x ≠ -3`

Do `x ∈ Z => x + 5` và `x + 3∈ Z`

Để `A ∈ Z <=> x + 5 ⋮x + 3`

`<=> x + 3 + 2 ⋮ x + 3`

Do `x + 3 ⋮ x + 3`

Nên `2 ⋮ x + 3`

`=> x + 3 ∈ Ư(2) =` {`-2;-1;1;2`}

`=> x ∈` {`-5;-4;-2;-1`} (Thỏa mãn)

Vậy ...

------------------------------

`B =(x-2)/(x+1)`

Điều kiện: `x ≠ -1`

Do `x ∈ Z => x -2` và `x + 1 ∈ Z`

Để `B ∈ Z <=> x -2 ⋮x + 1`

`<=> x + 1 - 3 ⋮x + 1`

Do `x + 1 ⋮x + 1`

Nên `3⋮x + 1`

`=> x + 1 ∈ Ư(3) =` {`-3;-1;1;3`}

`=> x ∈` {`-4;-2;0;2`} (Thỏa mãn)

Vậy ...

\(A=\dfrac{x+5}{x+3}\in Z\)

\(\Rightarrow\left(x+5\right)⋮\left(x+3\right)\)

Mà \(\left(x+3\right)⋮\left(x+3\right)\)

\(\Rightarrow\left(x+5\right)-\left(x+3\right)⋮\left(x+3\right)\)

\(\Rightarrow2⋮\left(x+3\right)\)

\(\Rightarrow\left(x+3\right)\inƯ\left(2\right)\)

\(\Rightarrow\left(x+3\right)\in\left\{1;-1;-2;2\right\}\)

Ta có bảng giá trị:

Vậy \(x\in\left\{-2;-4;-1;-5\right\}\)

Những câu còn lại, cách làm tương tự, nếu như còn thắc mắc thì bạn tag mình nhé.

d: \(\left|-5-\sqrt{2}\right|=5+\sqrt{2}\)

c: \(\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

d: \(\left|-\dfrac{4}{15}\right|=\dfrac{4}{15}\)

a: \(\left|3,02\right|=3,02\)

a: \(\left|-\dfrac{1}{3}\right|-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=\dfrac{1}{3}-1+\dfrac{1}{4}:2=-\dfrac{2}{3}+\dfrac{1}{8}=\dfrac{-16}{24}+\dfrac{3}{24}=-\dfrac{13}{24}\)

b: \(\left(\dfrac{2}{3}\right)^3+\sqrt{\dfrac{49}{81}}-\left|-\dfrac{7}{3}\right|:3\)

\(=\dfrac{8}{27}+\dfrac{7}{9}-\dfrac{7}{3}\cdot\dfrac{1}{3}\)

\(=\dfrac{8}{27}+\dfrac{7}{9}-\dfrac{7}{9}=\dfrac{8}{27}\)

c: \(\sqrt{\dfrac{25}{49}}+\left(5555\right)^0+\left|-\dfrac{2}{7}\right|\)

\(=\dfrac{5}{7}+1+\dfrac{2}{7}\)

=1+1=2

a: \(\sqrt{50}>\sqrt{49}\)

mà \(\sqrt{49}=7\)

nên \(\sqrt{50}>7\)

b: \(\sqrt{27}>\sqrt{25}=5\)

=>\(\dfrac{4}{\sqrt{27}}< \dfrac{4}{5}\)

c: \(\dfrac{3}{\sqrt{7}}>1;\dfrac{\sqrt{7}}{3}< 1\)

Do đó: \(\dfrac{3}{\sqrt{7}}>\dfrac{\sqrt{7}}{3}\)

Bài 2:

a:

\(-4,4\left(9\right)-5,8\left(1\right)\simeq-4,5-5,8=-10,3\)

 \(-4,4\left(9\right)-5,8\left(1\right)\)

\(=-\dfrac{9}{2}-\dfrac{-523}{90}=-\dfrac{9}{2}+\dfrac{523}{90}=\dfrac{118}{90}=\dfrac{59}{45}\)

b:

\(-12,\left(7\right)\cdot3,\left(12\right)\simeq-12,8\cdot3,1\simeq-40\)

 \(-12,\left(7\right)\cdot3,\left(12\right)\)

\(=-\dfrac{115}{9}\cdot\dfrac{103}{33}=\dfrac{11845}{297}\)

c: \(9,\left(49\right):\left[-5,\left(09\right)\right]\simeq9,5:\left(-5,1\right)\simeq-1,9\)

\(9,\left(49\right):\left[-5,\left(09\right)\right]\)

\(=\dfrac{940}{99}:\dfrac{-56}{11}=\dfrac{940}{99}\cdot\dfrac{11}{-56}\)

\(=\dfrac{940}{-56}\cdot\dfrac{1}{9}=-\dfrac{235}{14\cdot9}=-\dfrac{235}{126}\)

Bài 1:

a: \(9,4\simeq9\)

b: \(3,51\simeq4\)

c: \(-7,505\simeq-8\)

d: \(-1.199\simeq-1\)

Sửa đề: \(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{2^{12}\cdot9^6+8\cdot9^5}\)

\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^{12}+2^3\cdot3^{10}}\)

\(=\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^3\cdot3^{10}\left(2^9\cdot3^2+1\right)}\)

\(=\dfrac{2^9}{3^6}\cdot\dfrac{2}{1028\cdot9+1}=\dfrac{2^{10}}{729\left(1028\cdot9+1\right)}\)

sorry mình ấn nhầm nha

\(B=\left(-2\right)+\left(-2\right)^2+...+\left(-2\right)^{2024}\)

=>\(\left(-2\right)\cdot B=\left(-2\right)^2+\left(-2\right)^3+...+\left(-2\right)^{2025}\)

=>\(-2B-B=\left(-2\right)^2+\left(-2\right)^3+...+\left(-2\right)^{2025}-\left(-2\right)-\left(-2\right)^2-...-\left(-2\right)^{2024}\)

=>\(-3B=-2^{2025}+2\)

=>\(B=\dfrac{-2^{2025}+2}{-3}=\dfrac{2^{2025}-2}{3}\)

a: Xét ΔAHI vuông tại H và ΔAKI vuông tại K có

AI chung

\(\widehat{HAI}=\widehat{KAI}\)

Do đó: ΔAHI=ΔAKI

b: ΔAHI=ΔAKI

=>IH=IK

Xét ΔIBC có

IM là đường cao

IM là đường trung tuyến

Do đó: ΔIBC cân tại I

=>IB=IC

Xét ΔIHB vuông tại H và ΔIKC vuông tại K có

IB=IC

IH=IK

Do đó: ΔIHB=ΔIKC

=>BH=CK

Do \(f\left(3\right)=f\left(-3\right)\Rightarrow a.3^2+b.3+c=a.\left(-3\right)^2+b.\left(-3\right)+c\)

\(\Rightarrow9a+3b+c=9a-3b+c\)

\(\Rightarrow6b=0\)

\(\Rightarrow b=0\)

\(\Rightarrow f\left(x\right)=ax^2+c\)

\(f\left(-x\right)=a.\left(-x\right)^2+x=ax^2+c\)

\(\Rightarrow f\left(x\right)=f\left(-x\right)\)