a/

\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)

\(\Rightarrow x=12\)

\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)

\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)

Vậy x = 2023

\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)

=>Trong 2 số phải có 1 số âm và 1 số dương

Mà \(2-x>\dfrac{4}{5}-x\)

=>\(\dfrac{4}{5}< x< 2\)

Vậy...

Lời giải:

Điều kiện: $x\neq -2; x\neq -2; x\neq -8; x\neq -14$

Đề bài 

$\Rightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}$

$\Rightarrow 12=x$ (thỏa mãn)

1: =>|1/4x^2+1/45|=1/20

=>1/4x^2+1/45=1/20 hoặc 1/4x^2+1/45=-1/20

=>1/4x^2=1/36

=>x^2=1/36:1/4=1/9

=>x=1/3 hoặc x=-1/3

2: =(x^2-3)(x^2-2x)

=x(x-2)(x^2-3)

a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)

=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)

=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)

=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)

=> \(x=\dfrac{-1}{11}\)

Đây toán 8 mà? :v

a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)

\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)

\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)

\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)

\(\Leftrightarrow\left(11+1\right)x=0\)

\(\Leftrightarrow11x+1=0;x=0\)

\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)

Vậy....

bai nay de\

Bạn coi lại xem có viết nhầm chỗ nào trong biểu thức không? Biểu thức này nội việc rút gọn thôi đã "xấu" rồi.

Bài 1 :

a) \(\sqrt{4\left(a-3\right)^2}+2\sqrt{\left(a^2+4a+4\right)}\)

= \(2\left|a-3\right|+2\left|a+2\right|\)

\(=2.\left(-a+3\right)+2\left(-a-2\right)\)

b) có sai đề ko ?

c) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)

ĐKXĐ: x>=0; x<>1

a: \(B=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left(\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\right)\)

\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left[\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2\right]\)

\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

b: Khi x=4-2căn 3=(căn 3-1)^2 thì \(B=\dfrac{\sqrt{3}-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}\)

c: B=2/3

=>căn x/căn x+1=2/3

=>căn x=2

=>x=4

d: \(B-1=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{x}+1}< 0\)

=>B<1

e: B>1

=>-1/căn x+1>0

=>căn x+1<0(vô lý)

=>KO có x thỏa mãn

f: B nguyên khi căn x chia hết cho căn x+1

=>căn x+1-1 chia hết cho căn x+1

=>căn x+1=1 hoặc căn x+1=-1(loại)

=>căn x=0

=>x=0