b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:

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

\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)

\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)

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: Cho từng cái < hoặc > 0 rồi giải ra tìm điều kiện của x

Bài 2:

Phân tích số 12 ra là:

3 x 4 = 12

-3 x (-4) = 12

Ta thấy: 

3 + 4 = 7

-3 + (-4) = -7 (đáp ứng đúng yêu cầu đề)

=> a = -3 và b = -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 )

=> \(0,75< 1,02< 3\)

bài 1/ 

a) ta có: \(A=\frac{15}{x-1}\)

Để A là phân số \(\Rightarrow x-1\ne0\)

                          \(\Rightarrow x\ne1\)

b) Nếu x = 7

\(\Rightarrow A=\frac{15}{7-1}\)

\(\Rightarrow A=\frac{15}{6}\)

Nếu x = -3

\(\Rightarrow A=\frac{15}{-3-1}\)

\(\Rightarrow A=\frac{15}{-4}\)

Nếu x = 4

\(\Rightarrow A=\frac{15}{4-1}\)

\(\Rightarrow A=\frac{15}{3}=5\)

c) Ta có: \(B=5\)

\(\Leftrightarrow A=\frac{15}{x-1}=5\)

\(\Leftrightarrow x-1=3\)

\(\Leftrightarrow x=4\)

Bài 2/

a) \(\frac{x}{3}=\frac{2}{6}\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\)

b) \(-\frac{x}{14}=\frac{10}{-7}\)

\(\Leftrightarrow7x=140\)

\(\Leftrightarrow x=20\)

hok tốt!!

Bài 1:

a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)

b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)

Bài 2:

 \(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)

Bài 3:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)

Bài 2: 

a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)

\(\Leftrightarrow x-2=4-x\)

\(\Leftrightarrow2x=6\)

hay x=3

b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)

\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)

\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)

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)

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

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

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

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 \(\left(x;y\right)=\left(2;-3\right)\)