tìm các số nguyên x để các biểu thức sau có giá trị là một số nguyên
A=x+5/x+3
B=x-2/x+1
C=x+1/x-3
D=2x-5/2x+1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{1}{5}.\left(x+2\right)^2+\dfrac{1}{3}.\left(2x-2\right)^3=\dfrac{1}{5}.\left(x+2\right)^2+\dfrac{1}{3}.2^3\)
\(\Rightarrow\left(2x-2\right)^3=2^3\)
\(\Rightarrow2x-2=2\)
\(\Rightarrow2x=2+2\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4\div2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
`1/5 . (x+2)^2 + 1/3 . (2x - 2)^3 = 1/5 . (x+2)^2 + 1/3 . 2^3`
`<=> 1/5 . (x+2)^2 - 1/5 . (x+2)^2+ 1/3 . (2x - 2)^3 = 1/3 . 2^3`
`<=> 0 + 1/3 . (2x - 2)^3 = 1/3 . 2^3`
`<=> 1/3 . (2x - 2)^3 = 1/3 . 2^3`
`<=> 1/3 : 1/3 . (2x - 2)^3 = 2^3`
`<=> 1 . (2x - 2)^3 = 2^3`
`<=> (2x - 2)^3 = 2^3`
`<=> 2x - 2 = 2`
`<=> 2x = 2+2 `
`<=> 2x = 4`
`<=> x = 4 : 2`
`<=> x = 2`
Vậy `x = 2`
\(4\dfrac{3}{8}+5\dfrac{2}{3}\)
\(=4+5+\dfrac{3}{8}+\dfrac{2}{3}\)
\(=9+\dfrac{9}{24}+\dfrac{16}{24}\)
\(=9+\dfrac{25}{24}\)
\(=10\dfrac{1}{24}\)
\(4\dfrac{3}{8}+5\dfrac{2}{3}\)
\(=\dfrac{35}{8}+\dfrac{17}{3}\)
\(=\dfrac{105}{24}+\dfrac{136}{24}\)
\(=\dfrac{241}{24}\)
Bài 4:
a: \(216x^3+27y^3=27\left(8x^3+y^3\right)\)
\(=27\left[\left(2x\right)^3+y^3\right]\)
\(=27\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
b: \(64a^3-8=8\left(8a^3-1\right)\)
\(=8\left[\left(2a\right)^3-1^3\right]\)
\(=8\left(2a-1\right)\left(4a^2+2a+1\right)\)
c: \(x^3+8=x^3+2^3=\left(x+2\right)\left(x^2-2x+4\right)\)
d: \(27x^3-8y^3=\left(3x\right)^3-\left(2y\right)^3\)
\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x\cdot2y+\left(2y\right)^2\right]\)
\(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
Bài 5:
a: \(3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)
\(=3\left(x^2-2xy+y^2\right)-2\left(x^2+2xy+y^2\right)-\left(x^2-y^2\right)\)
\(=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\)
\(=2y^2-10xy\)
b: \(\left(x-y\right)^3-3\left(x-y\right)^2\cdot x+3\left(x-y\right)\cdot x^2-x^3\)
\(=\left(x-y-x\right)^3\)
\(=\left(-y\right)^3=-y^3\)
c: \(\left(3x+3\right)^3-2\left(x+1\right)^3-\left(5x-1\right)^2\)
\(=27\left(x+1\right)^3-2\left(x+1\right)^3-\left(5x-1\right)^2\)
\(=25\left(x+1\right)^3-25x^2+10x-1\)
\(=25x^3+75x^2+75x+25-25x^2+10x-1\)
\(=25x^3+50x^2+85x+24\)
d: \(\left(-2x+3\right)^3-\left(x+1\right)^3+\left(3x-1\right)^2\)
\(=\left(-2x+3-x-1\right)\left[\left(-2x+3\right)^2+\left(-2x+3\right)\left(x+1\right)+\left(x+1\right)^2\right]+\left(3x-1\right)^2\)
\(=\left(-3x+2\right)\left(4x^2-12x+9-2x^2+x+3+x^2+2x+1\right)+\left(3x-1\right)^2\)
\(=\left(-3x+2\right)\left(3x^2-9x+13\right)+\left(3x-1\right)^2\)
\(=-9x^3+27x^2-39x+6x^2-18x+26+9x^2-6x+1\)
\(=-9x^3+42x^2-63x+27\)
\(\left(x^2+1\right)\left(x-1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-1=0
=>x=1
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: \(\left|3,02\right|=3,02\)
`180 = 2.2. 3.3 . 5`
`2024 = 2.2.2 . 11 . 23`
`1500 = 2.2.3.5.5.5`
`400 = 2.2.2.2.5.5`
`504 = 2.2.2.3.3.7`
`890 = 2.5.89`
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= (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.