e: vecto AM=(x-3;y+1)

vecto BM=(x+1;y-2)

vecto AC=(-2;0)

vecto AM=2*vecto BM-3*vecto AC

=>x-3=2*(x+1)+6 và y+1=2(y-2)

=>x-3=2x+8 và y+1=2y-4

=>x=-11 và y=5

f: Tọa độ H là:

\(\left\{{}\begin{matrix}x=\dfrac{3-1+1}{3}=1\\y=\dfrac{-1+2-1}{3}=0\end{matrix}\right.\)

g: K thuộc Oy nên K(0;y)

vecto AB=(-4;3)

vecto AK=(-3;y+1)

A,K,B thẳng hàng

=>\(-\dfrac{3}{-4}=\dfrac{y+1}{3}\)

=>y+1=9/4

=>y=5/4

h: P thuộc Ox nên P(x;0)

vecto PA=(x-3;1)

vecto PC=(x-1;1)

ΔPAC vuông tại P

=>vecto PA*vecto PC=0

=>(x-3)(x-1)+1=0

=>x^2-4x+3+1=0

=>x=2

=>P(2;0)

đáp số : A O C D E H T I K M N U W V X S B Đ Y

a: Ta có: \(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}+\dfrac{4}{3}\sqrt{3}+3\sqrt{6}+6\sqrt{3}\)

\(=\dfrac{37}{3}\sqrt{3}+3\sqrt{6}\)

c: Ta có: \(\left(\sqrt{12}+2\sqrt{27}\right)\cdot\dfrac{\sqrt{3}}{2}-\sqrt{150}\)

\(=\left(2\sqrt{3}+6\sqrt{3}\right)\cdot\dfrac{\sqrt{3}}{2}-5\sqrt{6}\)

\(=12-5\sqrt{6}\)

f: Thay x=0 và y=5 vào (d), ta được: 

m-1=5

hay m=6

e: Thay x=1 và y=4 vào (d),ta được:

2m+3+m-1=4

=>3m+2=4

hay m=2/3

\(A:SO_2\\ B : Fe_2O_3\\ D : SO_3\\ E : H_2O\\ F: H_2SO_4\\ G : CuSO_4\\ H : K_2SO_3\\ I : BaSO_3\\ K : KCl\\ L : BaSO_4 \\ M : HCl\)

\(4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2\\ 2SO_2 + O_2 \xrightarrow{t^o,V_2O_5} 2SO_3\\ SO_3 + H_2O \to H_2SO_4 \\ 2H_2SO_4 + Cu \\ CuSO_4 + SO_2 + 2H_2O\\ SO_2 + 2KOH \to K_2SO_3 + H_2O\\ K_2SO_3 + BaCl_2 \to BaSO_3 + 2KCl\\ BaSO_3 + H_2SO_4 \to BaSO_4 + SO_2 + H_2O\\ SO_2 + Cl_2 + 2H_2O \to 2HCl + H_2SO_4\)

Em cung cấp thông tin của đề rõ ràng lại nhé :3

\(e,3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=8\left(x\ge0\right)\\ \Leftrightarrow\sqrt{x}\left(3\sqrt{2}-5\sqrt{8}+7\sqrt{18}\right)=8\\ \Leftrightarrow\sqrt{x}\left(3\sqrt{2}-10\sqrt{2}+21\sqrt{2}\right)=8\\ \Leftrightarrow14\sqrt{2x}=8\Leftrightarrow\sqrt{2x}=\dfrac{4}{7}\Leftrightarrow2x=\dfrac{16}{49}\Leftrightarrow x=\dfrac{8}{49}\left(tm\right)\)

\(f,\sqrt{4x+20}-\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\left(x\ge-5\right)\\ \Leftrightarrow2\sqrt{x+5}-\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x+5}=4\\ \Leftrightarrow0\sqrt{x+5}=4\\ \Leftrightarrow\sqrt{x+5}=0\Leftrightarrow x+5=0\Leftrightarrow x=-5\left(tm\right)\)

e) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}=8\left(đk:x\ge0\right)\)

\(\Leftrightarrow3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=8\)

\(\Leftrightarrow14\sqrt{2x}=8\Leftrightarrow\sqrt{2x}=\dfrac{8}{14}\Leftrightarrow2x=\dfrac{16}{49}\Leftrightarrow x=\dfrac{8}{49}\left(tm\right)\)

f) \(\sqrt{4x+20}-\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)

\(\Leftrightarrow2\sqrt{x+5}-\sqrt{x+5}-\sqrt{x+5}=4\)

\(\Leftrightarrow0=4\left(VLý\right)\)

Vậy \(x\in\left\{\varnothing\right\}\)

d) \(x^2=a\left(a\ge0\right)\)

\(\Rightarrow x=\sqrt{a}\)

e) \(x^2=\dfrac{4}{9}\)

\(\Rightarrow x^2=\left(\pm\dfrac{2}{3}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

f) \(x^2-\dfrac{16}{25}=0\)

\(\Rightarrow x^2=\dfrac{16}{25}\)

\(\Rightarrow x^2=\left(\pm\dfrac{4}{5}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

g) \(x^2-\dfrac{7}{36}=0\)

\(\Rightarrow x^2=\dfrac{7}{36}\)

\(\Rightarrow x^2=\left(\pm\sqrt{\dfrac{7}{36}}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{7}{36}}\\x=-\sqrt{\dfrac{7}{36}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{7}}{6}\\x=-\dfrac{\sqrt{7}}{6}\end{matrix}\right.\)

h) Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2+1\ge1>0\forall x\)

mà \(x^2+1=0\)

nên không tìm được giá trị nào của x thoả mãn đề bài.

\(e,\left(x-2\right)^2-16=0\\ \Leftrightarrow\left(x-6\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\\ f,x^2-5x-14=0\\ \Leftrightarrow\left(x-7\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ g,8x\left(x-3\right)+x-3=0\\ \Leftrightarrow\left(8x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\\x=3\end{matrix}\right.\)

e)\(\left(x-2-4\right)\left(x-2+4\right)=\left(x-6\right)\left(x+2\right)\)

f)\(x^2-5x-14=x^2-2.\dfrac{5}{2}x+\dfrac{25}{2}+\dfrac{3}{2}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{2}\)

e: Ta có: \(\left(x-2\right)^2-16=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)

f: Ta có: \(x^2-5x-14=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\)