Giúp mk câu b với ạ
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\(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)
\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+3.105\)
\(\Leftrightarrow-6x+93=175x+455\)
\(\Leftrightarrow181x=-362\)
\(\Leftrightarrow x=-2\)
\(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)\(\Leftrightarrow\dfrac{84x+63}{105}-\dfrac{90x-30}{105}=\dfrac{175x+140}{105}+\dfrac{315}{105}\)
\(\Leftrightarrow84x+63-90x+30=175x+140+315\)
\(\Leftrightarrow84x-90x-175x=140+315-63-30\)
\(\Leftrightarrow-181x=362\)
\(\Leftrightarrow x=-2\)
8 How much do these apples cost?
9 THis is a blue car
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13 How wide if the Great wall
1 send - will receive
2 do - will improve
3 find - will give
4 will go - has
5 will go- gets
6 doesn't phone - will leave
7 don't study - won't pass
8 rains - won't have to
9 won't be able - watch
10 can't move - isn't
11 study - will pass
1.
a, \(\left(C\right)x^2+y^2-6x-2y+6=0\)
\(\Leftrightarrow\left(C\right)\left(x-3\right)^2+\left(y-1\right)^2=4\)
\(\Rightarrow\) Tâm \(I=\left(3;1\right)\), bán kính \(R=2\)
b, Tiếp tuyến đi qua A có dạng: \(\left(\Delta\right)ax+by-5a-7b=0\left(a^2+b^2\ne0\right)\)
Ta có: \(d\left(I;\Delta\right)=\dfrac{\left|3a+b-5a-7b\right|}{\sqrt{a^2+b^2}}=2\)
\(\Leftrightarrow\left|a+3b\right|=\sqrt{a^2+b^2}\)
\(\Leftrightarrow6ab+8b^2=0\)
\(\Leftrightarrow2b\left(3a+4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b=0\\3a+4b=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\Delta_1:x=5\\\Delta_2:4x-3y+1=0\end{matrix}\right.\)
TH1: \(\Delta_1:x=5\)
Tiếp điểm có tọa độ là nghiệm hệ: \(\left\{{}\begin{matrix}x=5\\x^2+y^2-6x-2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y^2-2y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\Rightarrow\left(5;1\right)\)
TH2: \(\Delta_2:4x-3y+1=0\)
Tiếp điểm có tọa độ là nghiệm hệ: \(\left\{{}\begin{matrix}4x-3y+1=0\\x^2+y^2-6x-2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{5}\\y=\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left(\dfrac{7}{5};\dfrac{11}{5}\right)\)
Kết luận: Phương trình tiếp tuyến: \(\left\{{}\begin{matrix}\Delta_1:x=5\\\Delta_2:4x-3y+1=0\end{matrix}\right.\)
Tọa độ tiếp điểm: \(\left\{{}\begin{matrix}\left(5;1\right)\\\left(\dfrac{7}{5};\dfrac{11}{5}\right)\end{matrix}\right.\)
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Should Julia be helped with the sewing ?
1: \(P=\left(\dfrac{\sqrt{x-2}\left(3-\sqrt{x-2}\right)}{9-x+2}+\dfrac{x+7}{11-x}\right):\left(\dfrac{3\sqrt{x-2}+1-\sqrt{x-2}+3}{\sqrt{x-2}\left(\sqrt{x-2}-3\right)}\right)\)
\(=\dfrac{3\sqrt{x-2}-x+2+x+7}{11-x}:\dfrac{2\sqrt{x-2}+4}{\sqrt{x-2}\left(\sqrt{x-2}-3\right)}\)
\(=\dfrac{3\sqrt{x-2}+9}{11-x}\cdot\dfrac{\sqrt{x-2}\left(\sqrt{x-2}-3\right)}{2\sqrt{x-2}+4}\)
\(=\dfrac{-3\left(\sqrt{x-2}+3\right)}{2\left(\sqrt{x-2}+2\right)}\cdot\dfrac{\sqrt{x-2}}{\sqrt{x-2}+3}\)
\(=\dfrac{-3\sqrt{x-2}}{2\sqrt{x-2}+4}\)
2: Đặt căn x-2=a(a>=0)
=>P=-3a/(2a+4)
P nguyên
=>-3a chia hết cho 2a+4
=>-6a chia hết cho 2a+4
=>-6a-12+12 chia hết cho 2a+4
=>2a+4 thuộc {1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}
=>a thuộc {1;4}
=>x-2=1 hoặc x-2=16
=>x=3 hoặc x=18