\(\frac{2}{2a-1}.\sqrt{5x^4\left(1-4a+4a^2\right)}\)

\(=\frac{2}{2a-1}.\sqrt{5x^4\left(2a-1\right)^2}\)

\(=\frac{2}{2a-1}.x^2.\left(2a-1\right).\sqrt{5}\)

\(=2\sqrt{5}x^2\)

Gọi tam giác cân là ABC (cân tại A), đường cao AH.
Gọi cạnh đáy của tam giác cân là a, cạnh bên là b. Theo đề bài:
10a = 12b
=> a/b = 6/5
Đặt a = 6k, b = 5k
Xét tam giác AHC vuông tại H:
AH^2 + HC^2 = AC^2
<=> 10^2 + a^2/4 = b^2
<=> a^2/4 = b^2 - 100
<=> (6k)^2/4 = (5k)^2 - 100
<=> 9k^2 = 25k^2 - 100
<=> 16k^2 = 100 <=> k = 10/4
=> a = 6k = 6.10/4 = 15 (cm)
=> S_ABC = 1/2BC.AH = 1/2a.10 = 5a = 5.15 = 75 (cm^2)

75 cm vuông bạn nhé

\(\frac{3x}{\sqrt{3x+10}}=\sqrt{3x+1-1}\)

<=> 3x\(=\sqrt{3x+1-1}.\sqrt{3x+10}\)

<=> (3x)² = (\(\sqrt{3x+1-1}.\sqrt{3x+10}\))²

<=> 9x² = 9x² + 30

<=> x = 0

=> x = 0

Đặt ẩn phụ

\(a,\sqrt[3]{x+1}=x+1\)

\(\Leftrightarrow\left(x+1\right)=\left(x+1\right)^3\)

\(\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+1+1\right)\left(x+1-1\right)=0\)

\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)

\(\Leftrightarrow x=0\left(h\right)x=-1\left(h\right)x=-2\)

Từ NC = 3 NA => NC = ³/₄ CA

Kẻ NH _|_CD

=> NH // AD

Theo Ta-let có

\(\frac{NH}{AD}=\frac{CN}{CA}=\frac{\frac{3}{4}CA}{CA}=\frac{3}{4}\)

\(\Rightarrow NH=\frac{3AD}{4}=\frac{3.4}{4}=3\)

Theo Pytago có \(AD^2+DC^2=AC^2\)

               \(\Leftrightarrow4^2+8^2=AC^2\)

              \(\Leftrightarrow AC^2=80\)

                \(\Leftrightarrow AC=4\sqrt{5}\)

                \(\Rightarrow NC=\frac{3}{4}AC=\frac{3}{4}.4\sqrt{5}=3\sqrt{5}\)

Áp dụng định lí Pytago \(NH^2+HC^2=NC^2\)

                                  \(\Leftrightarrow3^2+HC^2=45\)

                                \(\Leftrightarrow HC^2=36\)

                                 \(\Leftrightarrow HC=6\)

CÓ \(MC=\frac{CD}{2}=\frac{8}{2}=4\)

\(\Rightarrow HM=HC-CM=6-4=2\)

Áp dụng Pytago

\(HN^2+HM^2=NM^2\)

\(\Leftrightarrow3^2+2^2=NM^2\)

\(\Leftrightarrow MN^2=13\)

\(\Leftrightarrow MN=\sqrt{13}\)

ĐKXĐ: \(x\ge0\)

\(Pt\Leftrightarrow\left(\sqrt{x+3}-\sqrt{2x+2}\right)+\left(\sqrt{3x+1}-2\sqrt{x}\right)=0\)

   \(\Leftrightarrow\frac{x+3-2x-2}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{3x+1-4x}{\sqrt{3x+1}+2\sqrt{x}}=0\)

\(\Leftrightarrow\frac{1-x}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{1-x}{\sqrt{3x+1}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(1-x\right)\left(\frac{1}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{1}{\sqrt{3x+1}+2\sqrt{x}}\right)=0\)

Dễ thấy cái ngoặc to > 0 nên x = 1 (thỏa mãn)

\(\frac{\sqrt{x^2+a^2}+a}{a-\sqrt{x^2-a^2}}=\frac{\left(a+\sqrt{x^2+a^2}\right)\left(a+\sqrt{x^2-a^2}\right)}{\left(a-\sqrt{x^2-a^2}\right)\left(a+\sqrt{x^2-a^2}\right)}=\frac{...}{a^2-x^2+a^2}=\frac{...}{2a^2-x^2}\)

\(\left(2\sqrt{1+a}\right)^2=4\left(1+a\right)=\left(\sqrt{1+x}+\sqrt{1+y}\right)^2\le2\left(x+y+2\right)\)

\(\Leftrightarrow\)\(x+y\ge2a\)

Áp dụng bđt Bunyakovsky: \(\left(\sqrt{1+x}+\sqrt{1+y}\right)^2\le2\left(x+y+2\right)\)

\(\Rightarrow4\left(a+1\right)\le2\left(x+y+2\right)\Leftrightarrow4a\le2\left(x+y\right)\Leftrightarrow x+y\ge2a\)

\(2x+3+\sqrt{4x^2+9x+2}=2\sqrt{x+2}+\sqrt{4x+1}\left(x\ge-\frac{1}{4}\right)\)

\(\Leftrightarrow2\left(x+2\right)-1+\sqrt{\left(x+2\right)\left(4x+1\right)}=2\sqrt{x+2}+\sqrt{4x+1}\)

\(\Leftrightarrow4\left(x+2\right)-2+2\sqrt{x+2}.\sqrt{4x+1}=4\sqrt{x+2}+2\sqrt{4x+1}\)

Đặt \(\hept{\begin{cases}2\sqrt{x+2}=a\left(a\ge0\right)\\\sqrt{4x+1}=b\left(b\ge0\right)\end{cases}\Rightarrow}a^2-b^2=4\left(x+2\right)-4x-1=7\)\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=7\)(1)

\(pt:a^2-2+ab=2a+2b\)

\(\Leftrightarrow a\left(a+b\right)-2\left(a+b\right)=2\)

\(\Leftrightarrow\left(a-2\right)\left(a+b\right)=2\)(2)

Nhân chéo 2 vế của (1) với (2) được

\(7\left(a-2\right)\left(a+b\right)=2\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow7\left(a-2\right)=2\left(a-b\right)\left(Do\left(a+b\right)>0\right)\)

\(\Leftrightarrow7a-14=2a-2b\)

\(\Leftrightarrow5a=14-2b\)

\(\Leftrightarrow10\sqrt{x+2}=14-2\sqrt{4x+1}\)

\(\Leftrightarrow5\sqrt{x+2}=7-\sqrt{4x+1}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{4x+1}\le7\\25\left(x+2\right)=49-14\sqrt{4x+1}+4x+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}0\le4x+1\le49\\21x=-14\sqrt{4x+1}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le0\\441x^2=196\left(4x+1\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le0\\441x^2-784x-196=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-\frac{1}{4}\le x\le0\\49\left(9x+2\right)\left(x-2\right)=0\end{cases}}\)

\(\Leftrightarrow x=-\frac{2}{9}\left(TmĐKXĐ\right)\)

Vậy

Incursion_03 em thử nha, sai thì thôi ạ, em hơi nghiện liên hợp r.

ĐK: x>=-1/4

PT \(\Leftrightarrow2x+\frac{31}{9}+\sqrt{4x^2+9x+2}-\frac{4}{9}=2\sqrt{x+2}-\frac{8}{3}+\sqrt{4x+1}-\frac{1}{3}+3\)

\(\Leftrightarrow2\left(x+\frac{2}{9}\right)+\frac{\left(x+\frac{2}{9}\right)\left(4x+\frac{73}{9}\right)}{\sqrt{4x^2+9x+2}+\frac{4}{9}}=\frac{4\left(x+\frac{2}{9}\right)}{2\sqrt{x+2}+\frac{8}{3}}+\frac{4\left(x+\frac{2}{9}\right)}{\sqrt{4x+1}+\frac{1}{3}}\)

\(\Leftrightarrow\left(x+\frac{2}{9}\right)\left[2+\frac{4x+\frac{73}{9}}{\sqrt{4x^2+9x+2}+\frac{4}{9}}-4\left(\frac{1}{2\sqrt{x+2}+\frac{8}{3}}+\frac{1}{\sqrt{4x+1}+\frac{1}{3}}\right)\right]=0\)

Cái ngoặc to em chịu:( đang suy nghĩ