Chương II : Số nguyên

là sao,đề của bạn ảo quá mk ko hiểu

????;12<200

10<200

-200<200

vậy.ak.kakaka

\(2x^2-1=49\)

\(2x^2=49+1\)

\(2x^2=50\)

\(x^2=50:2\)

\(x^2=25\)

\(\Rightarrow\left\{{}\begin{matrix}x=5^2\\x=\left(-5\right)^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)\(\left(TM\right)\)

Vậy \(x\in\left\{5;-5\right\}\) là giá trị cần tìm

\(2x^2-1=49\)

\(2x^2=49+1\)

\(2x^2=50\)

\(x^2=25\)

\(x={5;-5}\)

p/s : đăng câu này r mà má

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

\(A=-a-b+c+a+b+c\)

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

\(A=2c\)

Thay 1;-1 ;-2 vào a;b;c ta có:

\(A=\left(-1+1-2\right)+\left(1+1+2\right)\)

\(A=-2+4=2\)

\(2x^2-1=49\)

\(2x^2=49+1\)

\(2x^2=50\)

\(x^2=50:2\)

\(x^2=25\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=5^2\\x^2=\left(-5\right)^2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\) \(\left(TM\right)\)

Vậy \(x\in\left\{5;-5\right\}\) là giá trị cần tìm

\(2x^2-1=49\)

\(2x^2=49+1\)

\(2x^2=50\)

\(x^2=25\)

\(x={5;-5}\)

2 x ² - 1 = 49

2 x ² = 49 + 1

2 x ² = 50

x ² = 25

x = 5 : - 5

\(2x+12=3\left(x-7\right)\)

\(\Leftrightarrow2x+12=3x-21\)

\(\Leftrightarrow-x+12=-21\)

\(\Leftrightarrow-x=-33\)

\(\Leftrightarrow x=33\)

Vậy \(x=33\)

\(2x+12=3(x-7)\)

\(2x+12=3x-21\)

\(2x=3x-21-12\)

\(2x=3x-33\)
\(3x-2x=33\)

\(x=33\)

a) 5-|x+1| = 29

|x+1| = 5 - 29

|x+1| = -24

\(\Rightarrow\left[{}\begin{matrix}x+1=24\\x+1=-24\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=23\\x=-25\end{matrix}\right.\)

Vậy x = -25 hoặc 23

a) \(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}+\dfrac{1}{195}\)

\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-.....+\dfrac{1}{13}-\dfrac{1}{15}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{15}\right)\)

\(=\dfrac{1}{2}.\dfrac{4}{15}=\dfrac{2}{15}\)

b) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)

\(=\dfrac{4}{9}.\left(-7\right)+\dfrac{59}{9}\left(-7\right)\)

\(=-7\left(\dfrac{4}{9}+\dfrac{59}{9}\right)\)

\(=-7.7=-49\)

c) \(\left(3\dfrac{2}{5}-2\dfrac{2}{5}\right).\left(-\dfrac{5}{3}\right)+3.\left(2\dfrac{1}{2}:\dfrac{1}{2}\right)\)

\(=\left(\dfrac{17}{5}-\dfrac{12}{5}\right).\left(-\dfrac{5}{3}\right)+3.5\)

\(=-\dfrac{5}{3}+15=13\dfrac{1}{3}\)

d) \(1\dfrac{13}{5}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}+1\dfrac{19}{60}\right):1\dfrac{23}{24}\)

\(=\dfrac{2}{7}+78\dfrac{8}{15}:\dfrac{47}{24}\)

( bạn tự tính nốt câu này nha ! )

*)Nếu x\(\ge\)1 ta có:

3(x-1)-6x=0

<=>3x-3-6x=0

<=>-3x=3

<=>x=-1(L)

*)Nếu 1>x\(\ge\)0 ta có:

3(1-x)-6x=0

<=>3-3x-6x=0

<=>3-9x=0

<=>9x=3

<=>x=\(\dfrac{1}{3}\)(TM)

*)Nếu x<0 ta có:

3(1-x)+6x=0

<=>3-3x+6x=0

<=>3+3x=0

<=>3x=-3

<=>x=-1(TM)

Vậy tập nghiệm S={\(\dfrac{1}{3};-1\)}

đéo hiểu kiểu gì ?