Ta có :

\(<=> (x-1)(2-y)-(x-1)(y-2) = 0 \\ <=> (x-1)[2-y-(y-2)] = 0 \\ <=> (x-1)[2-y-y+2] = 0 \\ <=> (x-1)(-2y+4) = 0 \\ => \Bigg[ \begin{matrix} x-1=0\ (1)\\ -2y+4=0\ (2)\\ \end{matrix}\\ Ta\ có :\ (1) <=> x=1\\ Ta\ có :\ (2) <=> -2y=-4 <=> y = 2\\ Vậy\ x = 1,\ y=2. \)

Đó là bài giải, cảm ơn bạn đã cho câu hỏi khó đó!

Chúc bạn học tốt!

a) Số hs trung bình là:

40 . 35% = 14 (hs)

Số hs khá là:

(40-14) . 8/13 = 16 (hs)

Số hs giỏi là:

40 - (14+16) = 10 (hs)

b) Tỉ số % của số hs khá so vs số hs của lớp là

16 : 40 . 100% = 40%

Đáp số a) 16hs khá

                10 hs giỏi

            b) 40%

Số học sinh giỏi là:

40 . 25% = 10 (học sinh)

Số học sinh khá là: 

40 . 3/5 = 24 (học sinh

Số học sinh trung bình là:

40 - 10 - 24 = 6 (học sinh)

2.x+10/3=22/3

2.x=22/3-10/3

2.x=4

x=4:2

x=2

vậy x=2

Trả lời:

\(2x+\frac{31}{3}=7\frac{1}{3}\)

\(\Leftrightarrow2x+\frac{31}{3}=\frac{22}{3}\)

\(\Leftrightarrow2x=\frac{-9}{3}\)

\(\Leftrightarrow2x=-3\)

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

Vậy \(x=\frac{-3}{2}\)

16/15+-5/3

=-3/5

gạch dọc chữ e nha 

cảm ơn các bạn trước nhá

mình thêm nữa là cách trình bày của câu này như thế nào:

x chia hất cho 12, x chia hết cho 25, x chia hết cho 30 và 0< x<500

Ư(90)={1;2;3;5;6;9;10;15;18;30;45;90;}

Ư(126)={1;126;63;2;3;42;6;21;7;18;14;9}

Ư(84)={1;2;3;4;6;7;12;14;21;28;42;84}

Ư(63)={1;3;7;9;21;63}

Ư(105)={1;3;5;;7;15;21;35;105}

mik chỉ biết làm tới đây thôi ! xin lỗi nha

1.It's a boarding school

-> Is it a boarding school?

2.Students study and live here.

-> Do students study and live here?

3.The school has students from all over Australia and overseas

->Does the school have students from all over Australia and overseas?

4.The school is surrounded by moutains and green fields

-> Is the school surrounded by moutains and green fields?

5.Students learn English with English spealing teachers

-> Do students learn English with English spealing teachers?

6.We join many interesting clubs after school.

-> Do we/you join many interesting clubs after school?

7. Giống câu 8

8.My friend does drawings and paintings in the art club

-> Does my friend do drawings and paintings in the art club?

9.I am excited about the first day of school.

-> Am I excited about the first day of school?

10.Hung and Dung are excited about the first day of school.

-> Are Hung and Dung excited about the first day of school?

1.Rewrite the sentences in interrogative form

1.It's a boarding school

.......Is it a boarding school?..........................

2.Students study and live here.

...Who are study and live here?.........

3.The school has students from all over Australia and overseas

Where does the student come from?

4.The school is surrounded by moutains and green fields

.What is the thing surrounding the school?...

5.Students learn English with English spealing teachers

 Students learn English with whom ?

6.We join many interesting clubs after school.

What do they do after school?

7.My friend do drawings and paintings in the art club

What is your friend doing in the art club?

8.My friend does drawings and paintings in the art club

What is your friend doing in the art club?

9.I am excited about the first day of school.

How did you feel the first day of school?

10.Hung and Dung are excited about the first day of school.

What Did Hung and Dung feel on the first day of school?

* Hoặc có thể làm theo cách của @THTN .

Ta có \(\frac{2x+1}{x-3}=\frac{2x-6+7}{x-3}=\frac{2\left(x-3\right)+7}{x-3}=2+\frac{7}{x+3}\)

Vì \(2\inℤ\Rightarrow C\inℤ\Leftrightarrow\frac{7}{x-3}\inℤ\)

=> \(7⋮x-3\)

=> \(x-3\inƯ\left(7\right)\)

=> \(x-3\in\left\{-1;-7;1;7\right\}\)

=> \(x\in\left\{2;-4;4;10\right\}\)

Vậy C\(\inℤ\Leftrightarrow x\in\left\{2;-4;4;10\right\}\)

\(C=\frac{2x+1}{x-3}=\frac{2\left(x-3\right)+7}{x-3}=2+\frac{7}{x-3}\)

Để C nguyên => \(\frac{7}{x-3}\)nguyên 

=> \(7⋮x-3\)

=> \(x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Vậy x thuộc các giá trị trên 

\(M=\left\{x\inℕ,n\inℕ^∗/n^2=x\right\}\)

\(B=\frac{x-5}{x+2}=\frac{x+2-7}{x+2}=1-\frac{7}{x+2}\)

Để B nguyên => \(\frac{7}{x+2}\)nguyên 

=> \(7⋮x+2\)

=> \(x+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Vậy x thuộc các giá trị trên 

Ta có \(\frac{x-5}{x+2}=\frac{x+2-7}{x+2}=1-\frac{7}{x+2}\)

=> \(B\inℤ\Leftrightarrow1-\frac{7}{x+2}\inℤ\)

Vì \(1\inℤ\Rightarrow B\inℤ\Leftrightarrow\frac{-7}{x+2}\inℤ\)

=> \(-7⋮x+2\)

=> \(x+2\inƯ\left(-7\right)\)

=> \(x+2\in\left\{1;7;-1;-7\right\}\)

=> \(x\in\left\{-1;5;-3;-9\right\}\)

Vậy với \(x\in\left\{-1;5;-3;-9\right\}\)thì B có giá trị nguyên