Bài 2: Cộng, trừ số hữu tỉ

Bài này khó quá :<< t ko biết làm

a)trời, nó dễ đến hiển nhiên luôn ý
x. số j ko cần biết mà = 0
thì 1 trong hai x hoặc căn x-2 sẽ là 0
căn mà ra 0 thì chỉ có căn 0 thôi
x-2=0 => x=2

hoặc x = 0
Từ đó:
2. căn 2-2 = 0
2. 0 = 0

b)cái này y chang cái trên, 1 trong 2 là 0
x căn để đc 0 thì chỉ có 0
còn x mũ 2 trừ 4 để bằng 0 thì x mũ 2 chỉ có 4

x mũ 2 = 4
x = 2
Đáp án: 2 hoặc 0

c) x phần x mà để đc bằng nhau thì |x| = x
đáp án: ∞

a, \(x.\sqrt{x-2}=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{x-2}=0\Rightarrow x-2=0^2\Rightarrow x-2=0\Rightarrow x=2\end{matrix}\right.\)

KL: Vậy \(x\in\left\{0;2\right\}\)

b, \(\sqrt{x}.\left(x^2-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\Rightarrow x=0^2\Rightarrow x=0\\x^2-4=0\Rightarrow x^2=4\Rightarrow x=\pm2\end{matrix}\right.\)

KL: Vậy \(x\in\left\{0;\pm2\right\}\)

Câu c mk ko biết lm

\(\dfrac{6}{7}:\left(\dfrac{1}{3}-\dfrac{-5}{2}\right)^2\)

= \(\dfrac{6}{7}:\left(\dfrac{17}{6}\right)^2\)

= \(\dfrac{216}{2023}\)

Ta có:

\(\dfrac{-13}{38}< 1\\ \dfrac{29}{-88}=\dfrac{-29}{88}< 1\\ \Rightarrow\dfrac{-13}{38}< \dfrac{29}{-88}\)

a. \(x=\dfrac{-2}{15}-\dfrac{3}{10}\)

\(x=\dfrac{-13}{30}\)

b. \(\dfrac{-11}{12}x+\dfrac{1}{4}=\dfrac{5}{6}\)

\(\dfrac{-11}{12}x=\dfrac{5}{6}-\dfrac{1}{4}\)

\(\dfrac{-11}{12}x=\dfrac{7}{12}\)

\(x=\dfrac{7}{12}:\dfrac{-11}{12}\)

\(x=\dfrac{-7}{11}\)

c:=>2003x+2005003=2004

=>2003x=-2002999

hay \(x\in\varnothing\)

Chu vi tam giác là 405x3=1215(cm)

Chu vi tứ giác là 310x4=1240(cm)

=>Chu vi hình tam giác kém hơn 25cm

a: \(\Leftrightarrow\left(\dfrac{3}{4}\right)^x=\left(\dfrac{4}{3}\right)^4=\left(\dfrac{3}{4}\right)^{-4}\)

=>x=-4

b: =>(x-2)(x-1)(x-3)=0

hay \(x\in\left\{2;1;3\right\}\)

c: =>(x-2)(x+3)=0

=>x=2 hoặc x=-3

a,\(x:\left(\dfrac{-3}{5}\right)^2=\dfrac{-3}{5}\)

\(\Leftrightarrow x=\dfrac{-3}{5}.\dfrac{9}{25}\Leftrightarrow x=\dfrac{-27}{125}\)

b,\(\left(\dfrac{-1}{3}\right)^3.x=\dfrac{1}{81}\)

\(\Leftrightarrow x=.\dfrac{1}{81}:\left(\dfrac{-1}{27}\right)\Leftrightarrow x=\dfrac{-1}{3}\)

c,(2x²)=16\(\Leftrightarrow\)x²=8\(\Leftrightarrow\)x=\(,\sqrt{8}\)

Giải:

a. \(x:\left(\dfrac{-3}{5}\right)^2=\dfrac{-3}{5}\)

\(\Rightarrow x=\left(\dfrac{-3}{5}\right)^2.\dfrac{-3}{5}\)

\(\Rightarrow x=\left(\dfrac{-3}{5}\right)^3\)

\(\Rightarrow x=\dfrac{-27}{125}\)

Vậy.................

b.\(\left(\dfrac{-1}{3}\right)^3.x=\dfrac{1}{81}\)

\(\Rightarrow x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)^3\)

\(\Rightarrow x=\dfrac{1}{81}:\dfrac{-1}{27}\)

\(\Rightarrow x=\dfrac{1}{81}.\dfrac{-27}{1}\)

\(\Rightarrow x=\dfrac{1}{3}.\dfrac{-1}{1}\)

\(\Rightarrow x=\dfrac{-1}{3}\)

vậy................

c.\(\left(2x^2\right)=16\)

\(\Rightarrow2x=\sqrt{16}=4\)

\(\Rightarrow x=4:2\)

\(\Rightarrow x=2\)

vậy....................

1.

a)\(-49+\left(-\dfrac{5}{6}\right)-\dfrac{17}{4}\)

\(=-49-\dfrac{5}{6}-\dfrac{17}{4}\)

\(=\dfrac{-588}{12}-\dfrac{10}{12}-\dfrac{51}{12}\)

\(=\dfrac{-588-10-51}{12}\)

\(=-\dfrac{649}{12}\)

b) \(5\dfrac{1}{2}+\left(-3\right)\)

\(=\dfrac{11}{2}-3\)

\(=\dfrac{11}{2}-\dfrac{6}{2}\)

\(=\dfrac{11-6}{2}\)

\(=\dfrac{5}{2}\)

c) \(4\dfrac{9}{11}+\left(2-2\dfrac{1}{11}\right)\)

\(=\dfrac{53}{11}+2-\dfrac{23}{11}\)

\(=\dfrac{53-23}{11}+2\)

\(=\dfrac{30}{11}+2\)

\(=\dfrac{30}{11}+\dfrac{22}{11}\)

\(=\dfrac{30+22}{11}\)

\(=\dfrac{52}{11}\)

2.

a) \(4,3-1,2=3,1\)

b) \(0-\left(-0,4\right)=0+0,4=0,4\)

c) \(-\dfrac{2}{3}-\dfrac{-1}{3}=-\dfrac{2}{3}+\dfrac{1}{3}=-\dfrac{1}{3}\)

d) \(-\dfrac{1}{2}-\dfrac{-1}{6}=-\dfrac{1}{2}+\dfrac{1}{6}=-\dfrac{3}{6}+\dfrac{1}{6}=-\dfrac{2}{6}=-\dfrac{1}{3}\)

\(A=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)

\(A=\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)

\(A=-2-0-\dfrac{1}{2}\)

\(A=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)

A = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

A = \(6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)

A = \(\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)

A = \(\left(-2\right)-0+\left(-\dfrac{1}{2}\right)\)

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

A = \(-\dfrac{5}{2}\)

\(A=\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)

\(A=\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)

\(A=-2-0-\dfrac{1}{2}\)

\(A=-2-\dfrac{1}{2}\)

\(A=-\dfrac{5}{2}\)

a: 2/3=12/18

-3/8=-12/32

6=-12/2

b: 2/3=16/24

-3/8=-9/24

6=144/24