| x + 1 | = | -3x = 2 |
mn giúp e với ạ, e đang cần gấp
cảm ơn ạ
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\(E=2.3+3.4+4.5+3.6+2.7+4.15=2\left(3+7\right)+3\left(4+6\right)+4\left(5+15\right)=2.10+3.10+4.20=20+30+80=130\)
\(F=3\left(12+13+14+15\right)+3\left(8+7+6+5\right)=3\left(12+8+13+7+14+6+15+5\right)=3\left(20+20+20+20\right)=3.80=240\)
f: Ta có: \(E=3\cdot\left(12+13+14+15\right)+3\left(8+7+6+5\right)\)
\(=3\left(12+13+14+15+8+7+6+5\right)\)
\(=3\cdot80=240\)
\(1,7x-8=4x+7\)
\(\Leftrightarrow7x-8-4x=7\)
\(\Leftrightarrow7x-4x=7+8\)
\(\Leftrightarrow3x=15\)
\(\Rightarrow x=5\)
\(2,3-2x=3\left(x+1\right)-x-2\)
\(\Leftrightarrow3-2x=2x+1\)
\(\Leftrightarrow-2x+3=2x+1\)
\(\Leftrightarrow-2x-2x=1-3\)
\(\Leftrightarrow-4x=-2\)
\(\Rightarrow x=\dfrac{1}{2}\)
\(3,5\left(3x+2\right)=4x+1\)
\(\Leftrightarrow5.3x+5.2=4x+1\)
\(\Leftrightarrow15x+10=4x+1\)
\(\Leftrightarrow15x-4x=1-10\)
\(\Leftrightarrow11x=-9\)
\(\Rightarrow x=\dfrac{-9}{11}\)
7. How often does he go to the library?
8. ....is the shortest student in his class.
9......is her address?
10... is shorter than Ba.
PT có 2 nghiệm `<=> \Delta' >=0`
`<=> 4(2m+3)^2 -4(4m^2-3) >=0`
`<=>16m^2+48m+36-16m^2+12>=0`
`<=>m >= -1`
Viet: `{(x_1+x_2=-2m-3),(x_1x_2=4m^2-3):}`
Theo đề: `x_1^2+x_2^2=1/2`
`<=>(x_1+x_2)^2-2x_1x_2=1/2`
`<=>(-2m-3)^2 -2(4m^2-3)=1/2`
`<=>-4m^2+12m+15=1/2`
`<=>` \(\left[{}\begin{matrix}m=\dfrac{6+\sqrt{94}}{4}\left(TM\right)\\m=\dfrac{6-\sqrt{94}}{4}\left(L\right)\end{matrix}\right.\)
Vậy....
ĐKXĐ: \(x\notin\left\{0;-9\right\}\)
Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)
Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)
\(\Leftrightarrow9x^2+81x+180=0\)
\(\Leftrightarrow x^2+9x+20=0\)
\(\Leftrightarrow x^2+4x+5x+20=0\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-4;-5}
VIII
1 Have you ever been
2 Did you go
3 went
4 got up
5 have never got
6 got
7 Have you ever seen
D
I
1 B
2 A
3 D
4 A
5 C
3x.(x-2)-x2+2x=0
⇔3x2-6x-x2+2x=0
⇔2x2-4x=0
⇔2x(x-2)=0
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
vậy x=0 và x=2
3x(x-2)-x^2+2x=0
<=>3x(x-2)-x(x-2)=0
<=>(3x-x)(x-2)=0
<=>2x(x-2)=0
<=>2x=0 hoặc x-2=0
<=>x=0 hoặc x=2
Hướng làm:
Thấy cả tử mẫu cộng lại đều bằng 2021 → Cộng thêm 1 rồi quy đồng với mỗi phân thức
\(\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\\ \Leftrightarrow\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\\ \Leftrightarrow\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}\right)=0\\ \Leftrightarrow x+2021=0\Leftrightarrow x=-2021\)
\(< =>\dfrac{x+2}{2019}+1+\dfrac{x+3}{2018}+1=\dfrac{x+4}{2017}+1+\dfrac{x}{2021}+1\)
\(< =>\dfrac{x+2+2019}{2019}+\dfrac{x+3+2018}{2018}=\dfrac{x+4+2017}{2017}+\dfrac{x+2021}{2021}\)
\(< =>\dfrac{x+2021}{2019}+\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}-\dfrac{x+2021}{2021}=0\)
\(< =>\left(x+2021\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2021}=\right)=0\)
\(< =>x+2021=0< =>x=-2021\)
Vậy....
Đề sai r bn