\(\Rightarrow\)\(2\left(\sqrt[3]{x}+1\right)=1\)\(\Rightarrow\sqrt[3]{x}+1=0,5\)

\(\Rightarrow\sqrt[3]{x}=-0,5\)\(\Rightarrow x=-0,125\)

N=\(\dfrac{x\sqrt{2}}{\sqrt{2x}\left(\sqrt{2}+\sqrt{x}\right)}+\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

N=\(\dfrac{\sqrt{x}}{\sqrt{2}+\sqrt{x}}+\dfrac{\sqrt{2}}{\sqrt{x}+\sqrt{2}}\)=1

a)=\(\dfrac{3\left(1+\sqrt{7}\right)}{\left(1-\sqrt{7}\right)\left(1+\sqrt{7}\right)}\)=\(\dfrac{3\left(1+\sqrt{7}\right)}{1-7}\)=\(\dfrac{3\left(1+\sqrt{7}\right)}{-6}\)=\(\dfrac{-1-\sqrt{7}}{2}\)

b)\(\dfrac{\left(3-\sqrt{8}\right)\left(\sqrt{8}-3\right)}{\left(\sqrt{8}+3\right)\left(\sqrt{8}-3\right)}\)=\(\dfrac{3\sqrt{8}-9-8+3\sqrt{8}}{8-9}\)=\(\dfrac{6\sqrt{8}-17}{-1}\)=\(17-6\sqrt{8}\)

chọn 1 nha

a) Ta có \(\widehat{N}=90^o-\widehat{Q}=90^o-60^o\)\(=30^o\)

Trong \(\Delta MNK\) vuông tại K

sinN=\(\dfrac{MK}{MN}\)\(\Rightarrow\)MN=\(\dfrac{MK}{sinN}\)=\(\dfrac{6}{sin30}\)=12

b) Trong \(\Delta MKQ\) vuông tại K có

KQ=KM.cotQ=6.cot60=3,5

a)A=\(\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

b) Thay x=3+2\(\sqrt{2}\)

A=\(\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}\)=\(\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2-2}}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)=\(\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}\)

A=\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)

c)Ta có \(\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)>0

\(\Rightarrow\dfrac{2}{\sqrt{x}}\)<1\(\Rightarrow\sqrt{x}\)>2\(\Rightarrow x>4\)

bạn tự vẽ hình giúp mik nha

kẻ đường cao AH (H\(\in\)BC)

trong \(\Delta ABH\) vuoonng tại H có

BH=AB.cosB=14.cos40=10,7(cm)

AH=\(\sqrt{AB^2-BH^2}\)(pytago)=\(\sqrt{14^2-10,7^2}\)=9(cm)

trong \(\Delta AHC\) vuông tại H có

HC=\(\sqrt{AC^2-HA^2}\)(pytago)=\(\sqrt{11^2-9^2}\)=6,3(cm)

mà: BC=BH+HC=10,7+6,3=17(cm)

bài 3

a) \(\sqrt{9x^2}=4\Rightarrow3x=4\)\(\Rightarrow\)\(x=\dfrac{4}{3}\)

b)\(\Rightarrow\)\(\left(x-\sqrt{5}\right)^2\)=0\(\Rightarrow x-\sqrt{5}=0\)

\(\Rightarrow x=\sqrt{5}\)

bài 2

a) ĐKXĐ: a\(\ge\)0, a\(\ne\)1

b)P=\(\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\).\(\dfrac{1+\sqrt{a}}{\sqrt{a}}\)

P=\(\dfrac{2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{1+\sqrt{a}}{\sqrt{a}}\)

P=\(\dfrac{2}{1-\sqrt{a}}\)

c) thay a=4 vào biểu thức ta có

P=\(\dfrac{2}{1-\sqrt{4}}\)=\(\dfrac{2}{1-2}\)=-2

d) để P=9 thì

\(\dfrac{2}{1-\sqrt{a}}=9\)\(\Rightarrow\)2=9(1-\(\sqrt{a}\))

\(\Rightarrow\)2=9-\(9\sqrt{a}\)\(\Rightarrow\)\(9\sqrt{a}=7\)\(\Rightarrow\)\(\sqrt{a}=\dfrac{7}{9}\)

\(\Rightarrow a=\dfrac{49}{81}\)

A=x-9-(x-\(4\sqrt{x}\)+4)=x-9-x+\(4\sqrt{x}\)-4=-13+\(4\sqrt{x}\)