Dài quá trôi hết đề khỏi màn hình: nhìn thấy câu nào giải cấu ấy

Bài 4:

\(A=\frac{\left(x-1\right)+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2}{\left(x+1\right)\left(x-1\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)

a) DK x khác +-1

b) \(dk\left(a\right)\Rightarrow A=\frac{2}{\left(x+1\right)}\)

c) x+1  phải thuộc Ước của 2=> x=(-3,-2,0))

1. a) Biểu thức a có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\end{cases}}\)

                                      \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)

                                       \(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)

   Vậy vs \(x\ne2,x\ne-2\) thì bt a có nghĩa

b)  \(A=\frac{x}{x+2}+\frac{4-2x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4-2x}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2-2x+4-2x}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)

\(=\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\)

 \(=\frac{x-2}{x+2}\)       

c) \(A=0\Leftrightarrow\frac{x-2}{x+2}=0\)             

\(\Leftrightarrow x-2=\left(x+2\right).0\)          

\(\Leftrightarrow x-2=0\)   

\(\Leftrightarrow x=2\)(ko thỏa mãn điều kiện )

=> ko có gía trị nào của x để A=0

1)

ĐKXĐ: x\(\ne\)3

ta có :

\(\frac{x^2-6x+9}{2x-6}=\frac{\left(x-3\right)^2}{2\left(x-3\right)}=\frac{x-3}{2}\)

để biểu thức A có giá trị = 1

thì :\(\frac{x-3}{2}\)=1

=>x-3 =2

=>x=5(thoả mãn điều kiện xác định)

vậy để biểu thức A có giá trị = 1 thì x=5

1)

\(A=\frac{x^2-6x+9}{2x-6}\)

A xác định

\(\Leftrightarrow2x-6\ne0\)

\(\Leftrightarrow2x\ne6\)

\(\Leftrightarrow x\ne3\)

Để A = 1

\(\Leftrightarrow x^2-6x+9=2x-6\)

\(\Leftrightarrow x^2-6x-2x=-6-9\)

\(\Leftrightarrow x^2-8x=-15\)

\(\Leftrightarrow x=3\) (loại vì không thỏa mãn ĐKXĐ)

a) Ta có: A = \(\left(\frac{x}{x-1}+\frac{x}{x^2-1}\right):\left(\frac{2}{x^2}-\frac{2-x^2}{x^3+x^2}\right)\)

A = \(\left(\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2\left(x+1\right)}{x^2\left(x+1\right)}-\frac{2-x^2}{x^2\left(x+1\right)}\right)\)

A = \(\left(\frac{x^2+x+x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x+2-2+x^2}{x^2\left(x+1\right)}\right)\)

A = \(\left(\frac{x^2+2x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x^2+2x}{x^2\left(x+1\right)}\right)\)

A = \(\frac{x\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x^2\left(x+1\right)}{x\left(x+2\right)}\)

A = \(\frac{x^2}{x+1}\)

b) ĐKXĐ: x \(\ne\)\(\pm\)1; x \(\ne\)0; x \(\ne\)-2

Ta có: A = 4

<=> \(\frac{x^2}{x+1}=4\)

<=> x² = 4(x + 1)

<=> x² - 4x - 4 = 0

<=>(x² - 4x + 4) - 8 = 0

<=> (x - 2)² = 8

<=> \(\orbr{\begin{cases}x-2=\sqrt{8}\\x-2=-\sqrt{8}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\sqrt{2}+2\\x=2-2\sqrt{2}\end{cases}}\)(tm)

Vậy ...

c) Ta có: A < 0

<=> \(\frac{x^2}{x+1}< 0\)

Do x² \(\ge\)0 => x + 1 < 0

=> x < -1

Vậy để A < 0 thì x < -1 và x khác -2

a) \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^{64}-1\right)\)

\(=\dfrac{3^{64}-1}{2}\)

b) \(\left(a+b+c\right)2+\left(a-b-c\right)2+\left(b-c-a\right)2+\left(c-a-b\right)2\)

\(=2\left[\left(a+b+c\right)+\left(a-b-c\right)+\left(b-c-a\right)+\left(c-a-b\right)\right]\)

\(=2\left(a+b+c+a-b-c+b-c-a+c-a-b\right)\)

\(=2.0\)

\(=0\)

c)\(\left(a+b+c+d\right)2+\left(a+b-c-d\right)2+\left(a+c-b-d\right)2+\left(a+d-b-c\right)2\)

\(=2\left(a+b+c+d+a+b-c-d+a+c-b-d+a+d-b-c\right)\)

\(=2.4a\)

\(=8a\)

bn ơi cho mk hỏi tại sao lại ko nhận 3 vậy !!!

a) Từ giả thiết : \(a^2+2c^2=3b^2+19\Rightarrow a^2+2c^2-3b^2=19\)

Ta có : \(\frac{a^2+7}{4}=\frac{b^2+6}{5}=\frac{c^2+3}{6}=\frac{3b^2+18}{15}=\frac{2c^2+6}{12}\)\(=\frac{a^2+7+2c^2+6-3b^2-18}{4+12-15}=\frac{14}{1}=14\)

\(\Rightarrow\)\(a^2=49\Rightarrow a=7\)

\(\Rightarrow\)\(b^2=64\Rightarrow b=8\)

\(\Rightarrow\)\(c^2=81\Rightarrow c=9\)

b) \(P=x^4+2x^3+3x^2+2x+1\)

\(=\left(x^4+2x^2+1\right)+\left(2x^3+2x\right)+x^2=\left(x^2+1\right)^2+2x\left(x^2+1\right)+x^2\)

\(=\left(x^2+x+1\right)^2\)

Vì \(x^2+x+1=\left(x^2+2x\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Nên \(P\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\)

Dấu bằng xảy ra khi và chỉ khi \(x=-\frac{1}{2}\)

Bố già giỏi qa

b: \(A=\frac{1-x}{2+x}-\frac{x-1}{x-2}+\frac{4-x^3}{4-x^2}\)

\(=\frac{-\left(x-1\right)}{x+2}-\frac{x-1}{x-2}+\frac{x^3-4}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{-\left(x-1\right)\left(x-2\right)-\left(x-1\right)\left(x+2\right)+x^3-4}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{-\left(x^2-3x+2\right)-\left(x^2+x-2\right)+x^3-4}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{-x^2+3x-2-x^2-x+2+x^3-4}{\left(x-2\right)\left(x+2\right)}=\frac{x^3-2x^2+2x-4}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2\left(x-2\right)+2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x^2+2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{x^2+2}{x+2}\)

d:

Sửa đề: Tìm x∈Z lớn nhất để A>0

A>0

=>\(\frac{x^2+2}{x+2}>0\)

=>x+2>0

=>x>-2

mà x là số nguyên lớn nhất có thể

nên x=-3

e: Để A là số nguyên thì \(x^2+2\) ⋮x+2

=>\(x^2-4+6\) ⋮x+2

=>6⋮x+2

=>x+2∈{1;-1;2;-2;3;-3;6;-6}

=>x∈{-1;-3;0;-4;1;-5;4;-8}

Bài giải

Cho:
[
A=\frac{1-x}{2+x}-\frac{x-1}{x-2}+\frac{4-x^3}{4-x^2},\quad (x\neq \pm2).
]

(b) Rút gọn A

Ta có:
[
\frac{1-x}{2+x}=-\frac{x-1}{x+2}.
]

Do đó:
[
-\frac{x-1}{x+2}-\frac{x-1}{x-2}=(x-1)\Big(-\frac{1}{x+2}-\frac{1}{x-2}\Big)
=-\frac{2x(x-1)}{x^2-4}.
]

Mặt khác:
[
\frac{4-x^3}{4-x^2}=\frac{x^3-4}{x^2-4}=x+\frac{4(x-1)}{x^2-4}.
]

Suy ra:
[
A=-\frac{2x(x-1)}{x^2-4}+x+\frac{4(x-1)}{x^2-4}
=x-\frac{2(x-1)(x-2)}{(x-2)(x+2)}.
]

[
\Rightarrow A=x-\frac{2(x-1)}{x+2}=\frac{x^2+2}{x+2}.
]

(d) Tìm (x\in \mathbb{Z}) nhỏ nhất để (A>0)

Vì (x^2+2>0) nên dấu của (A) phụ thuộc vào (x+2).
Điều kiện: (A>0 \iff x+2>0 \iff x>-2).
Số nguyên nhỏ nhất thoả mãn: (x=-1).

(e) Tìm (x\in\mathbb{Z}) để (A\in\mathbb{Z})

Ta có:
[
A=\frac{x^2+2}{x+2}=x-2+\frac{6}{x+2}.
]

Để (A\in\mathbb{Z}), cần (\dfrac{6}{x+2}\in \mathbb{Z}).
Vậy (x+2) phải là ước của 6: (\pm1,\pm2,\pm3,\pm6).

[
\Rightarrow x\in{-1,0,1,4,-3,-4,-5,-8}.
]

👉 Kết quả:

• (b) (A=\dfrac{x^2+2}{x+2},; x\neq \pm2).
• (d) (x=-1).
• (e) (x\in{-8,-5,-4,-3,-1,0,1,4}).

\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)

\(A=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3-3x}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(A=\frac{x^3-x^2+x-3-3x+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(A=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^3+1}\)