a) \(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}\)

\(=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}=-5\sqrt{x-1}\)

b) \(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)

\(=5\sqrt{y+4}+6\sqrt{y+4}-18\sqrt{y+4}=-7\sqrt{y+4}\)

c) \(P=\sqrt{y-2}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(=\sqrt{y-2}-24\sqrt{y-2}+28\sqrt{y-2}=5\sqrt{y-2}\)

a) \(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}.\)

\(M=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}\)

\(=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}\)

\(=-5\sqrt{x-1}\)

b) \(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)

\(N=\sqrt{25\left(y+4\right)}+\sqrt{36\left(y+4\right)}-2\sqrt{81\left(y+4\right)}\)

\(=5\sqrt{y+4}+6\sqrt{y+4}\)

\(=-7\sqrt{y+4}\)

c) \(P=\sqrt{\left(y-2\right)}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(P=\sqrt{\left(y-2\right)}-3\sqrt{64\left(y-2\right)}+4\sqrt{49\left(y-2\right)}\)

\(=\sqrt{y-2}-24\sqrt{y-2}+28\sqrt{y-2}\)

\(=5\sqrt{y-2}\)

a) \(A=2\sqrt{8}-3\sqrt{32}+\sqrt{50}\)

\(A=2\sqrt{4.2}-3\sqrt{16.2}+\sqrt{25.2}\)

\(A=2.2\sqrt{2}-3.4\sqrt{2}+5\sqrt{2}\)

\(A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}\)

\(A=\left(4-12+5\right)\sqrt{2}\)

\(A=-3\sqrt{2}\)

b) \(B=\sqrt{12}+4\sqrt{27}-3\sqrt{48}\)

\(B=\sqrt{4.3}+4\sqrt{9.3}-3\sqrt{16.3}\)

\(B=2\sqrt{3}+4.3\sqrt{3}-3.4\sqrt{3}\)

\(B=2\sqrt{3}\)

c) \(C=\sqrt{20a}+4\sqrt{45a}-2\sqrt{125a}\left(a\ge0\right)\)

\(C=\sqrt{4.5a}+4\sqrt{9.5a}-2\sqrt{25.5a}\)

\(C=2\sqrt{5a}+4.3\sqrt{5a}-2.5\sqrt{5a}\)

\(C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}\)

\(C=\left(2+12-10\right)\sqrt{5a}\)

\(C=4\sqrt{5a}\)

a) ta có \(2\sqrt{8}=2\sqrt{4.2}=4\sqrt{2},3\sqrt{32}=3\sqrt{16.2}=12\sqrt{2},\sqrt{50}=\sqrt{25.2}=5\sqrt{2}\)                               \(\Rightarrow A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}=-3\sqrt{2}\)                                                                                              b) ta có \(\sqrt{12}=\sqrt{4.3}=2\sqrt{3},4\sqrt{27}=4\sqrt{9.3}=12\sqrt{3},3\sqrt{48}=3\sqrt{16.3}=12\sqrt{3}\Rightarrow B=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}=26\sqrt{3}\)c) ta có \(\sqrt{20a}=\sqrt{4.5a}=2\sqrt{5a},4\sqrt{45a}=4\sqrt{9.5a}=12\sqrt{5a},2\sqrt{125a}=2\sqrt{25.5a}=10\sqrt{5a}\Rightarrow C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}=4\sqrt{5a}\)   

chịu.-.

HT~~~

a/ \(\sqrt{50a}=5\sqrt{2a}\)

b/ \(\sqrt{75x}=5\sqrt{3x}\)

a)\(5\sqrt{2a}\)

b)\(5\sqrt{3x}\)

Có hình vẽ : 

Dễ thấy S_ABCD = \(\frac{1}{2}\left(AH+CK\right).BD\)

mà lại có \(AH=AO.sin\alpha\) ; \(CK=OC.sin\alpha\)

=> S_ABCD = \(\frac{1}{2}\sin\alpha.AC.BD\)

Khi 2 đường chéo vuông góc với nhau thì 

\(H\equiv O\equiv K\Rightarrow AH=AO=CK\)

hay \(sin\alpha=1\)

Khi đó \(S_{ABCD}=\frac{1}{2}mn\)(đpcm) 

​Đặt AB = x , BC = x + 1 , AC = x + 2 , MH = a Xét 3 trường hợp 

Trường hợp 1 nếu góc B < 90o => BC > AC (khác đề)

Trường hợp 2 nếu góc B = 90 độ (khác đề)

Trường hợp 3 nếu góc B > 90o => AC > BC ( đúng) 

Nên ta sẽ đi xét trường hợp 3 : B > 90o ( bạn phải vẽ B > 90o nhé) HB = MH - BM

=> HB = a - (x+1)/2

=> HB^2 = (a - (x+1)/2)^2 HC = HB + BC

=> HC = a - x/2 + x

=> HC^2 = (a + (x+1)/2)^2 

Ta có AH^2 = AC^2 - HC^2 AH^2 = AB^2 - HB^2

 => AC^2 - HC^2 = AB^2 - HB^2 

<=> (x + 2)^2 - (a+ (x+1)/2)^2 = x^2 - (a - (x+1)/2)^2 

<=> x^2 - 4x - 4 - a^2 - ax - a - (x^2+2x+1)/4 = x^2 - a^2 + ax + a - (x^2+2x+1)/4 

<=> 2ax + 2a - 4x - 4 = 0 

<=> 2a(x+1) - 4(x+1) = 0 

<=> (x + 1).2(a - 2) = 0 

<=> x = -1 hoặc a = 2 hay AB = -1 hoặc HM = 2 

đi ngủ đê ae 

Không mất tính tổng quát ta chuẩn hóa \(AB=1\).

Dễ dàng suy ra \(AC=\sqrt{3},BC=2\).

\(AB+BM=AC+CM\)

\(\Leftrightarrow1+2-CM=\sqrt{3}+CM\)

\(\Leftrightarrow CM=\frac{3-\sqrt{3}}{2}\Rightarrow BM=\frac{1+\sqrt{3}}{2}\)

Kẻ \(AH\)vuông góc với \(BC\).

Suy ra \(BH=\frac{AB^2}{BC}=\frac{1^2}{2}=\frac{1}{2}\)

\(\Rightarrow MH=\frac{\sqrt{3}}{2}\)mà \(AH=\frac{AB.AC}{BC}=\frac{\sqrt{3}}{2}\)
suy ra \(MH=AH\)suy ra \(\Delta MAH\)vuông cân tại \(H\)

suy ra \(\widehat{AMH}=45^o\)

mà \(\widehat{AMH}=\widehat{ACM}+\widehat{CAM}\Leftrightarrow\widehat{CAM}=\widehat{AMH}-\widehat{ACM}=45^o-30^o=15^o\).

SOS hoặc SS đều ra.

nghĩa là gì ?