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\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

\(\Rightarrow x-1=\dfrac{6}{7}+\dfrac{1}{7}\times\left(\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{6}{7}+\dfrac{1}{7}\times1\\ \Rightarrow x-1=\dfrac{6}{7}+\dfrac{1}{7}=1\\ \Rightarrow x=1+1=2\)

= 2

\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)

Mai lam

a) Ta có:

1; 4; 7;...; 100 có (100 - 1) : 3 + 1 = 34 (số)

1 + 4 + 7+ ... + 100 = (100 + 1) × 34 : 2

= 101 × 17

(1 + 4 + 7 + ... + 100) : a = 17

101 × 17 : a = 17

a = 101 × 17 : 17

a = 100

b) (X - 1/2) × 5/3 = 7/4 - 1/2

(X - 1/2) × 5/3 = 5/4

X - 1/2 = 5/4 : 5/3

X - 1/2 = 3/4

X = 3/4 + 1/2

X = 5/4

a) (1 + 4 + 7 +...+ 100) : a = 17

1717 : a = 17

a = 101

b) \(\left(x-\dfrac{1}{2}\right)\times\dfrac{5}{3}=\dfrac{7}{4}-\dfrac{1}{2}\)

\(\left(x-\dfrac{1}{2}\right)\times\dfrac{5}{3}=\dfrac{10}{8}\)

\(\left(x-\dfrac{1}{2}\right)=\dfrac{10}{8}\div\dfrac{5}{3}\)

\(\left(x-\dfrac{1}{2}\right)=\dfrac{10}{8}\times\dfrac{3}{5}\)

\(\left(x-\dfrac{1}{2}\right)=\dfrac{3}{4}\)

\(x-\dfrac{1}{2}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}+\dfrac{1}{2}\)

\(x=\dfrac{5}{4}\)

a. (1+4+7+...+100):a=17 

=> ( 100 + 1) x 32 : 2 : a = 17

=> 1717 : a = 17 

=> a = 101

b. \(\left(X-\dfrac{1}{2}\right).\dfrac{5}{3}=\dfrac{7}{4}-\dfrac{2}{4}\)

\(\left(X-\dfrac{1}{2}\right).\dfrac{5}{3}=\dfrac{5}{4}\)

\(X-\dfrac{1}{2}=\dfrac{3}{4}\)

\(X=\dfrac{1}{4}\)

Bài 1:

a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)

\(=\dfrac{9}{20}-\dfrac{2}{9}\)

\(=\dfrac{41}{180}\)

b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)

\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)

\(=\dfrac{5}{6}\times\dfrac{12}{7}\)

\(=\dfrac{10}{7}\)

c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)

\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{7}{9}\times1\)

\(=\dfrac{7}{9}\)

Bài 2:

a) \(2\times\left(x-1\right)=4026\)

\(\left(x-1\right)=4026\div2\)

\(x-1=2013\)

\(x=2014\)

Vậy: \(x=2014\)

b) \(x\times3,7+6,3\times x=320\)

\(x\times\left(3,7+6,3\right)=320\)

\(x\times10=320\)

\(x=320\div10\)

\(x=32\)

Vậy: \(x=32\)

c) \(0,25\times3< 3< 1,02\)

\(\Leftrightarrow0,75< 3< 1,02\) ( S )

=> \(0,75< 1,02< 3\)

lop 1 moi la la do

Toán lớp 6 mới đúng

ai thấy đúng **** cho mk

\(a.\dfrac{1}{4}+\dfrac{3}{4}x=-0,3\)

\(\Leftrightarrow\dfrac{3}{4}x=-\dfrac{11}{20}\)

\(\Leftrightarrow x=-\dfrac{11}{15}\)

\(b.\dfrac{4}{7}x-\dfrac{7}{3}=\dfrac{1}{21}\)

\(\Leftrightarrow\dfrac{4}{7}x=\dfrac{50}{21}\)

\(\Leftrightarrow x=\dfrac{25}{6}\)

a: \(\dfrac{x-3}{5-x}=\dfrac{5}{7}\left(x\ne5\right)\)

=>7(x-3)=5(5-x)

=>7x-21=25-5x

=>12x=46

=>x=23/6

b: \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)(ĐKXĐ: \(x\notin\left\{1;-7\right\}\))

=>(x-2)(x+7)=(x+4)(x-1)

=>\(x^2+5x-14=x^2+3x-4\)

=>5x-14=3x-4

=>2x=10

=>x=5(nhận)