thank you bạn nhiều nha

thank

a) = x² -4x + 4+1 = (x-2)² +1

GYNN = 1

B) = (x² - 8x +16) + ( y² +4y +4) + 2016 -16 -4

     = (x - 4)² + ( y +2)² + 1996

GTNN = 1996

c) =(x-y)² +xy -x +y +1 hình như câu này bn chép sai đề nên k có GTNN

( ĐÂY là những bài khó , có thể bn k hiu nhưng đúng 100%, bn cứ chép ra lop thầy giảng cho)

câu c) có thể là y-1 mà bn nhầm y+1

a) 2-(x+3) = 1+2+3+...+99

1+2+3+...+99 → có 99 số hạng

2-(x+3) = (1+99).99 : 2 

2-(x+3) = 4950

x+3 = 2 + 4950 

x+3 = 4952

x = 4952 - 3

x = 4949

b) (x+1)+(x+2)+...+(x+100) = 5750

→ có 100 cặp

(x+x+x+...+x) + ( 1+2+3+...+100 ) = 5750

=> 100x + 5050 = 5750

100x = 5750 - 5050

100x = 700

x = 700 : 100

x = 7

 0o0 Nguyễn Đoàn Tuyết Vy 0o0 bà kêu tui học tốt có nghĩa là học giốt đúng ko

b)(x+1)+(x+2)+(x+3)+(x+4)+...............+(x+100)=5750

(x+x+x+x+...+x) + (1+2+3+4+...+100) = 5750

     100x             +           5050           = 5750

                       100x                          = 5750 - 5050

                       100x                          = 700

                         x                             = 700 : 100

                         x                             = 7

Vậy ...

\(a,x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)

\(b,x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x+\frac{43}{5}=\frac{37}{4}\)

=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)

\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)

=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)

=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)

d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)

=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)

=> \(\frac{98x}{198}=99\)

=>  98x = 99 . 198

=> 98x = 19602

=> x = 19602 : 98 = 9801/49

a) \(x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{71}{8}\)

b) \(x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x=\frac{37}{4}-\frac{61}{8}\)

=> \(x=\frac{13}{8}\)

c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)

=> \(x-\frac{61}{8}=3.\frac{1}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}\)

=> \(x=\frac{73}{8}\)

d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)

=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)

=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)

=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)

=> \(x.\frac{98}{99}=198\)

=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)

\(b,\text{ }x+7\frac{5}{8}=9\frac{1}{4}\)

\(x=9\frac{1}{4}-\left(7+\frac{5}{8}\right)\)

\(x=9+\frac{1}{4}-7-\frac{5}{8}\)

\(x=\left(9-7\right)+\left(\frac{1}{4}-\frac{5}{8}\right)\)

\(x=2+-\frac{3}{8}\)

\(x=2\frac{-3}{8}\)

a)th1:x+3-3x+2=(-2x)+5

th2:x+3+3x+2=4x+5

b)x=0

a,-12(x-5)+7(3-x)=20

-12x+60+21-7x=20

-19x=-61

x=\(\frac{61}{19}\)

b,30(x+1)-3(x-5)-15x=25

30x+30+15-3x-15x=25

12x=-20

x=\(-\frac{20}{12}\)

timx thuoc z ma

2^x+1-3=13

2^x+1=13+3

2^x+1=16

2^x+1=2⁴

Vậy x+1=4

x=4-1=3

Vậy x=3

a) 2x+1-3=13

2x+1=13+3

2x+1=16

2x+1=24

Vậy x+1=4

x=4-1=3

Vậy x=3

b) (x+1)³+3=30

(x+1)³ = 30 - 3

(x+1)³ = 27 = 3³

=> x+1 = 3

x = 3 - 1

x = 2

Vậy x = 2

TICK MK NHA

2\(^{x+1}\) - 3 = 13

\(\Rightarrow2^{x+1}=16\) \(\Rightarrow2x.2=16\) \(\Rightarrow\) 2x = 8 \(\Rightarrow\) x = 4

( x + 1 ) \(^3\) + 3 = 30

\(\Rightarrow\) \(\)\(\left(x+1\right)^3=27\)

3\(^3\) = 27

\(\Rightarrow x+1=3\Rightarrow x=2\)

a) Đặt \(\frac{x}{5}=\frac{y}{7}=k\)

\(\Rightarrow\hept{\begin{cases}x=5k\\y=7k\end{cases}}\)

\(\Rightarrow xy=5k.7k\)

\(\Rightarrow140=35k^2\)

\(\Rightarrow k^2=4\)

\(\Rightarrow\orbr{\begin{cases}k=2\\k=-2\end{cases}}\)

Với k = 2 ta có :

+) \(\frac{x}{5}=2\Rightarrow x=10\)

+) \(\frac{y}{7}=2\Rightarrow y=14\)

Với k = -2 ta có :

+) \(\frac{x}{5}=-2\Rightarrow x=-10\)

+) \(\frac{y}{7}=-2\Rightarrow y=-14\)

Vậy  \(\left(x;y\right)=\left\{\left(10;14\right);\left(-10;-14\right)\right\}\)

b) Ta có :

\(x:y:z\)\(=\)\(2:5:7\)\(\Rightarrow\)\(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\)\(\Rightarrow\)\(\frac{3x}{6}=\frac{2y}{10}=\frac{z}{7}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\frac{3x}{6}=\frac{2y}{10}=\frac{z}{7}=\frac{3x+2y-z}{6+10-7}=\frac{27}{9}=3\)

+) \(\frac{x}{2}=3\Rightarrow x=6\)

+) \(\frac{y}{5}=3\Rightarrow y=15\)

+) \(\frac{z}{7}=3\Rightarrow z=21\)

Vậy x = 6, y = 15 và z = 21

_Chúc bạn học tốt_

a, x.y/5.7=140/35

=140/35=4

x/5=4/7

x/7=5/4

x.7=5.4

x.7=20

x=20;7

x=20/7

b,chịu

tk thì tk ko tk cx đc

a, \(\frac{x}{5}=\frac{y}{7}\left(x.y=140\right)\)

Đặt \(\frac{x}{5}=\frac{y}{7}=k\)

\(\Rightarrow7x=5y\)

\(\Rightarrow x.y=7k.5k=35k^2=140\)

\(\Rightarrow k^2=4\Rightarrow k=\pm2\)

\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=2.7=14\\y=2.5=10\end{cases}}\\\hept{\begin{cases}x=\left(-2\right).7=-14\\y=\left(-2\right).5=-10\end{cases}}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\hept{\begin{cases}x=2.7=14\\y=2.5=10\end{cases}}\\\hept{\begin{cases}x=\left(-2\right).7=-14\\y=\left(-2\right).5=-10\end{cases}}\end{cases}}\)

Vậy ....

b, \(x:y:z=2:5:7\left(3x+2y-z=27\right)\)

Đặt \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=k\)

\(\Leftrightarrow x=2k;y=5k=z=7k\)

\(\Leftrightarrow3x+2y-z=6k+10k-7k=27\)

\(\Leftrightarrow x=6;y=15;z=21\)

Vậy ...