cho mình xem lại đề bài bạn nhé , mình thấy bạn làm không ổn

Bài tập mẫu `1:`

\(a,\left(3x+5\right)^2=\left(3x\right)^2+2\cdot3x\cdot5+5^2\\ =9x^2+30x+25\\ b,\left(2x-1\right)^2=\left(2x\right)^2-2\cdot2x\cdot1+1^2\\ =4x^2-4x+1\\ c,\left(3x+2y\right)^2=\left(3x\right)^2+3x\cdot2y\cdot2+\left(2y\right)^2\\ =9x^2+12xy+4y^2\)

Bài tập mẫu `2:`

\(a,\left(a+\dfrac{1}{4}\right)^2\\ =a^2+a\cdot2\cdot\dfrac{1}{2}+\dfrac{1}{2}\\ =a^2+a+\dfrac{1}{2}\\ b,\left(\dfrac{1}{2}x-\dfrac{2}{3}\right)^2\\ =\left(\dfrac{1}{2}x\right)^2-2\cdot\dfrac{1}{2}x\cdot\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2\\ =\dfrac{1}{4}x^2-\dfrac{2}{3}x+\dfrac{4}{9}\\ c,\left(\dfrac{1}{3}x+\dfrac{3}{2}y\right)^2\\ =\left(\dfrac{1}{3}x\right)^2+2\cdot\dfrac{1}{3}x\cdot\dfrac{3}{2}y+\left(\dfrac{3}{2}y\right)^2\\ =\dfrac{1}{9}x^2+xy+\dfrac{9}{4}y^2\)

Sửa `(a+b-c)/c=(a-b+c)/b=(-a+b+c)/a(a,,b,c\ne0)`

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

\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\left(a,b,c\ne0\right)\\ =\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}\\ =\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-c}{c}=1\\\dfrac{a-b+c}{b}=1\\\dfrac{-a+b+c}{a}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\left(1\right)\)

Thay `(1)` vào `M `ta được:

\(M=\dfrac{2a\cdot2b\cdot2c}{a\cdot b\cdot c}=8\)

Bài `2:`

`a)8x^2y^3(-1/2xyz^2)`

`=8*(-1/2)*(x^2*x)*(y^3*y)*z`

`=-4x^3y^4z`

`b)[12x^2y+(-6x^2y^3)+5xy]:3xy`

`=(12x^2y:3xy)-(6x^2y^3:3xy)+(5xy:3xy)`

`=4x-2xy^2+5/3`

`c)2x^2(3x^2-7xy+3y)`

`=(2x^2*3x^2)-(2x^2*7xy)+(2x^2*3y)`

`=6x^4-14x^3y+6x^2y`

`11` 

\(\dfrac{x+7}{x-7}-\dfrac{7}{x^2-7x}=\dfrac{1}{x}\left(ĐKXĐ:x\ne7,x\ne0\right)\\ \dfrac{x\left(x+7\right)}{x\left(x-7\right)}-\dfrac{7}{x\left(x-7\right)}=\dfrac{x-7}{x\left(x-7\right)}\\ x^2+7x-7=x-7\\ x^2+6x=0\\ x\left(x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\left(\text{loại}\right)\\x=-6\left(T/m\right)\end{matrix}\right.\)

`12`

\(\dfrac{x+5}{x}-\dfrac{x-7}{x+4}=\dfrac{x^2+35}{x^2+4x}\left(ĐKXĐ:x\ne0;x\ne-4\right)\\ \dfrac{\left(x+5\right)\left(x+4\right)}{x\left(x+4\right)}-\dfrac{x\left(x-7\right)}{x\left(x+4\right)}=\dfrac{x^2+35}{x\left(x+4\right)}\\ x^2+9x+20-x^2+7x=x^2+35\\ 16x-x^2-15=0\\ -\left(x^2-16x+15\right)=0\\ x^2-15x-x+15=0\\ x\left(x-15\right)-\left(x-15\right)=0\\ \left(x-15\right)\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-15=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\left(T/m\right)\\x=1\left(T/m\right)\end{matrix}\right.\)

Các câu còn lại làm tương tự

`b)`

+, Thay `x=-2` vào \(x^2+mx+m-3=0\) ta được :

`(-2)^2+m(-2)+m-3=0`

`4-2m+m-3=0`

`-m=-1`

`m=1`

Vậy với `x=-2` thì `m=1`

+, Thay `x=1` vào $x^2+mx+m-3=0$ ta được :

`1^2+m*1+m-3=0`

`1+m+m-3=0`

`2m=2`

`m=1`

Vậy `m=1` với `x=1`

toàn nam thanh nữ tú :)))

\(Sửa\dfrac{3}{4\cdot3}+\dfrac{3}{5\cdot4}+\dfrac{3}{5\cdot6}+...+\dfrac{3}{99\cdot100}\\ =3\cdot\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\right)\\ =3\cdot\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(\dfrac{1}{3}-\dfrac{1}{100}\right)=1-\dfrac{3}{100}\\ =\dfrac{97}{100}\)

bước 4 quên đổi dấu ạ