Câu này mình chưa học đến mình mới lớp 5 thôi đây toán lớp 7 chưa có ai chả lời được

Answer:

Câu 1:

\(5x+7y=40\)

\(\Rightarrow\hept{\begin{cases}5x=40\\7y=40\end{cases}}\Rightarrow\hept{\begin{cases}x=40:5\\y=40:7\end{cases}}\Rightarrow\hept{\begin{cases}x=8\\y=\frac{40}{7}\end{cases}}\)

Câu 2:

\(P=\frac{2x-5}{x+2}\left(x\ne-2\right)\)

\(=\frac{2x+4-9}{x+2}\)

\(=\frac{2x+4}{x+2}-\frac{9}{x+2}\)

\(=\frac{2\left(x+2\right)}{x+2}-\frac{9}{x+2}\)

\(=2-\frac{9}{x+2}\)

Mà để cho \(P\inℤ\) thì \(\frac{9}{x+2}\inℤ\)

\(\Rightarrow9⋮\left(x+2\right)\)

\(\Rightarrow x+2\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

Có bảng sau:

Vậy \(x\in\left\{-11;-5;-3;-1;1;7\right\}\) thì \(P\inℤ\)

Answer:

a) \(2.\left(\frac{-3}{2}\right)^2-\frac{7}{2}=2.\frac{9}{4}-\frac{7}{2}=1\)

b) \(5\frac{2}{3}x+1\frac{2}{3}=4\frac{1}{2}\)

\(\Rightarrow\frac{17}{3}x+\frac{5}{3}=\frac{9}{2}\)

\(\Rightarrow\frac{17}{3}x=\frac{17}{6}\)

\(\Rightarrow x=\frac{3}{6}=\frac{1}{2}\)

c) \(y=f\left(x\right)=5x^2+1\)

\(f\left(1\right)=5.1^2+1=5+1=6\)

\(f\left(-2\right)=5.\left(-2\right)^2+1=5+4+1=10\)

25-(-x+5)=-7

<=>-x+5=25+7

<=>-x+5=32

<=>-x=32+5

<=>-x=37

<=>x=-37

Vậy x=-37

-x + 5 = 25 - ( -7 )
-x +5 = 32
-x = 32 - 5 
-x = 27

Answer:

\(2-\frac{1}{3}\left|2x+1\right|=0,5\)

\(\Rightarrow2-\frac{1}{3}\left|2x+1\right|=\frac{1}{2}\)

\(\Rightarrow\frac{1}{3}\left|2x+1\right|=2-\frac{1}{2}\)

\(\Rightarrow\frac{1}{3}\left|2x+1\right|=\frac{3}{2}\)

\(\Rightarrow\left|2x+1\right|=\frac{9}{2}\)

\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{9}{2}\\2x+1=\frac{-9}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{7}{2}\\2x=\frac{-11}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=\frac{-11}{4}\end{cases}}\)

\(\text{Chứng minh:1+1=3}\)

\(\text{Ta có:6-6=9-9=0}\)

\(\text{mà:6-6=2x3-2x3}\)

\(\text{và:9-9=3x3-3x3}\)

\(\text{nên từ đó ta có:2x3-2x3=3x3-3x3}\)

\(\text{vậy ta có:2x(3-3)=3x(3-3)}\)

\(\text{Bỏ phép tính trong ngoặc ở 2 vế:}\)

\(\text{Ta còn:2=3}\)

\(\text{Vậy 1+1=2 thì 1+1=3}\)

\(\Rightarrow1+1=3\text{(dpcm)}\)

\(\text{Hok tốt!}\)

\(\text{@Kaito Kid}\)