a) Điều kiện xác định \(x\ge-2\)

Ta có \(\sqrt{x+2}-2x=3\)

\(\Leftrightarrow\sqrt{x+2}=3+2x\)\(\left(x\ge-\frac{3}{2}\right)\)

\(\Leftrightarrow\left(\sqrt{x+2}\right)^2=\left(2x+3\right)^2\)

\(\Leftrightarrow x+2=4x^2+12x+9\)

\(\Leftrightarrow4x^2+11x+7=0\)

\(\Leftrightarrow4x^2+4x+7x+7=0\)

\(\Leftrightarrow\left(x+1\right)\left(4x+7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-\frac{7}{4}\end{cases}}\)\(\Rightarrow x=-1\)( vì \(x\ge-\frac{3}{2}\)nên \(x\ne-\frac{7}{4}\))

b) Điều kiện xác định:  \(4-x^2\ge0\Rightarrow-2\le x\le2\)

\(2x-4\ge0\Rightarrow x\ge2\)

\(x\le2,x\ge2\)

nên xảy ra khi x=2

\(ĐKXĐ:x\ge-2\)

\(\sqrt{x+2}-2x=3\)

\(\Leftrightarrow x+2=\left(3+2x\right)^2\)

\(\Leftrightarrow x+2=9+12x+4x^2\)

\(\Leftrightarrow4x^2+11x+7=0\)

\(\Delta=11^2-4.7.4=9\)

\(\Leftrightarrow\orbr{\begin{cases}x_1=-1\left(TM\right)\\x_2=-\frac{7}{4}\left(TM\right)\end{cases}}\)

Vậy.........

hok tốt

7a có: \(\frac{1}{2}=x^2+y^2\ge\frac{\left(x+y\right)^2}{2}\)\(\Leftrightarrow x+y\le1\)

Áp dụng BD7 Cauchy-SChwarz 7a có: 

 \(V7=\frac{x}{y+1}+\frac{y}{x+1}=x-\frac{xy}{y+1}+y-\frac{xy}{x+1}\)

\(\le x+y-\frac{\left(x^2+y^2\right)}{2}\left(\frac{1}{y+1}+\frac{1}{x+1}\right)\)

\(\le1-\frac{\frac{1}{2}}{2}\cdot\frac{4}{1+2}=\frac{2}{3}=VP\)

Dấu "='' khi \(x=y=\frac{1}{4}\)

1

a) Ta có \(\frac{b^2-c^2}{\left(a+b\right).\left(a+c\right)}=\frac{\left(b+c\right)\left(b-c\right)}{\left(a+b\right).\left(a+c\right)}=\frac{\left(b+c\right)\left(a+b-a-c\right)}{\left(a+b\right).\left(a+c\right)}\)

\(=\frac{\left(b+c\right)\left(a+b\right)-\left(b+c\right).\left(a+c\right)}{\left(a+b\right).\left(a+c\right)}=\frac{b+c}{a+c}-\frac{b+c}{a+b}\)

Tương tự \(\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}=\frac{c+a}{b+a}-\frac{c+a}{b+c}\)

\(\frac{a^2-b^2}{\left(c+a\right).\left(c+b\right)}=\frac{a+b}{c+b}-\frac{a+b}{c+a}\)

Do đó \(\frac{b^2-c^2}{\left(a+b\right)\left(a+c\right)}+\frac{c^2-a^2}{\left(b+c\right)\left(b+a\right)}+\frac{a^2-b^2}{\left(c+a\right).\left(c+b\right)}\)

\(=\frac{b+c}{a+c}-\frac{b+c}{a+b}+\frac{c+a}{b+a}-\frac{c+a}{b+c}+\frac{a+b}{c+b}-\frac{a+b}{c+a}\)

\(=\frac{b+c-a-b}{a+c}+\frac{a+b-c-a}{b+c}+\frac{c+a-b-c}{a+b}\)

\(=\frac{c-a}{a+c}+\frac{b-c}{b+c}+\frac{a-b}{a+b}\)

1 b) Bạn có thể kham khảo ở đây https://h.vn/hoi-dap/tim-kiem?q=cho+x,y+th%E1%BB%8Fa+m%C3%A3n+:+[x+(c%C4%83n+x%5E2+2017)]nh%C3%A2n+[y++(c%C4%83n++y%5E2++2017)].+T%C3%ADnh+x+y&id=258448

\(_{S_{ABC}}=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\) với p=\(\frac{a+b+c}{2}\)

\(\Rightarrow\)S_ABC=84

Đặt BC=x \(\Rightarrow\)BH=x-16

\(\Rightarrow\)AB²=x(x-16) \(\Leftrightarrow\)15²=x(x-16) \(\Leftrightarrow\)x=25

\(\Rightarrow\)BC=25(cm),BH=25-16=9(cm)

AC=\(\sqrt{BC^2-AB^2}\)=20(cm)

AH=\(\sqrt{BH.HC}\)=12(cm

\(\cos30=\frac{\sqrt{3}}{2}\)