giải pt: \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)

làm thế này mà chả hiểu sao lại bị gạch, ai biết chỉ với, cảm ơn nak:

+ ĐK:\(\left\{{}\begin{matrix}x\ge1\\x+3-4\sqrt{x-1}\ge0\\x+8-6\sqrt{x-1}\ge0\end{matrix}\right.\) \(\Leftrightarrow x\ge1\)

+ pt đã cho \(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)

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

Th1: \(\left\{{}\begin{matrix}\sqrt{x-1}-2< 0\\\sqrt{x-1}-3< 0\end{matrix}\right.\)

(*) \(\Leftrightarrow2-\sqrt{x-1}+3-\sqrt{x-1}=1\Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\Leftrightarrow x=5\left(N\right)\)

Th2: \(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\ge0\end{matrix}\right.\)

(*) \(\Leftrightarrow\sqrt{x-1}-2+\sqrt{x-1}-3=1\Leftrightarrow2\sqrt{x-1}=6\Leftrightarrow\sqrt{x-1}=3\Leftrightarrow x=10\left(N\right)\)

Th3: \(\sqrt{x-1}-3< 0\le\sqrt{x-1}-2\)

(*) \(\Leftrightarrow\sqrt{x-1}-2+3-\sqrt{x-1}=1\Leftrightarrow1=1\left(đúng\right)\)

Kl: \(x\ge1\)

Đã bảo là liên hợp là ra mà đ tin hả Zũ ? -_- 

\(x^3+\sqrt{\left(x+1\right)^3}=9x+8\left(x\ge-1\right)\)

\(\Leftrightarrow\left(x^3+1\right)+\left(x+1\right)\sqrt{x+1}-9\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1+\sqrt{x+1}-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(Tm\right)\\x^2-x+\sqrt{x+1}-8=0\left(1\right)\end{cases}}\)

Giải \(\left(1\right)\Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)+\left(\sqrt{x+1}-2\right)=0\)

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

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

Vì  x > -1 nên dễ thấy cái ngoặc to > 0 

Do đó x = 3 

Vậy có 2 nghiệm -1 và 3 (nghiệm thứ 3 nào nữa nhỉ ? -,-'' )

+Tuấn 10B_2 (T ko biết đánh word nên dùng tạm .V)

GPT: \(\(\sqrt{x+3}+\sqrt[3]{x}=3\)\) (Bài này cách lp 9 dễ t ko giải nữa)

Vì \(\(f\left(x\right)=\sqrt{x+3}+\sqrt[3]{x}=3\)\) là hàm tăng trên tập [-3;\(\(+\infty\)\))

Ta có: Nếu \(\(x&gt;1\Leftrightarrow f\left(x\right)&gt;f\left(1\right)=3\)\)nên pt vô nghiệm

Nếu \(\(-3\le x&lt; 1\Leftrightarrow f\left(x\right)&lt; f\left(1\right)=3\)\)nên pt vô nghuêmj

Vậy x = 1

B2, GHPT: \(\(\hept{\begin{cases}2x^2+3=\left(4x^2-2yx^2\right)\sqrt{3-2y}+\frac{4x^2+1}{x}\\\sqrt{2-\sqrt{3-2y}}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\end{cases}}\)\)

ĐK \(\(\hept{\begin{cases}-\frac{1}{2}\le y\le\frac{3}{2}\\x\ne0\\x\ne-\frac{1}{2}\end{cases}}\)\)

Xét pt (1) \(\(\Leftrightarrow2x^2+3-4x-\frac{1}{x}=x^2\left(4-2y\right)\sqrt{3-2y}\)\)

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

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

Xét hàm số \(\(f\left(t\right)=t^3+t\)\)trên R có \(\(f'\left(t\right)=3t^2+1&gt;0\forall t\in R\)\)

Suy ra f(t) đồng biến trên R . Nên \(\(f\left(-\frac{1}{x}+1\right)=f\left(\sqrt{3-2y}\right)\Leftrightarrow-\frac{1}{x}+1=\sqrt{3-2y}\)\)

Thay vào (2) \(\(\sqrt{2-\left(1-\frac{1}{x}\right)}=\frac{\sqrt[3]{2x^2+x^3}+x+2}{2x+1}\)\)

\(\(\Leftrightarrow\sqrt{\frac{1}{x}+1}=\frac{\sqrt[3]{x^2\left(x+2\right)}+x+2}{2x+1}\)\)

\(\(\Leftrightarrow\left(2x+1\right)\sqrt{\frac{1}{x}+1}=x+2+\sqrt[3]{x^2\left(x+2\right)}\)\)

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

\(\(\Leftrightarrow f\left(\sqrt{1+\frac{1}{x}}\right)=f\left(\sqrt[3]{1+\frac{2}{x}}\right)\)\)

\(\(\Leftrightarrow\sqrt{1+\frac{1}{x}}=\sqrt[3]{1+\frac{2}{x}}\)\)

\(\(\Leftrightarrow\left(1+\frac{1}{x}\right)^3=\left(1+\frac{2}{x}\right)^2\)\)

Đặt \(\(\frac{1}{x}=a\)\)

\(\(\Rightarrow Pt:\left(a+1\right)^3=\left(2a+1\right)^2\)\)

Tự làm nốt , mai ra lớp t giảng lại cho ...

Giải hệ pt

1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)

2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)

3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)

4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)

m.n giúp e mấy bài này vs ạ!!

phương trình đã cho \(\Leftrightarrow\sqrt{2-\left(x-2\right)^2}+\sqrt{3-2\left(x-2\right)^2}=\sqrt{2}+\sqrt{3}\left(1\right)\)

Ta có \(2-\left(x-2\right)^2\le2\Rightarrow\sqrt{2-\left(x-2\right)^2}\le\sqrt{2}\)  tương tự ta có \(\sqrt{3-2\left(x-2\right)^2}\le\sqrt{3}\)      

\(\Rightarrow\sqrt{2-\left(x-2\right)^2}+\sqrt{3-2\left(x-2\right)^2}\le\sqrt{2}+\sqrt{3}\left(2\right)\)

Từ (1) và (2)=> dấu bằng xảy ra khi \(\left(x-2\right)^2=0\Leftrightarrow x=2\)

\(\left(\sqrt{8-\sqrt{X-3}}-\sqrt{5-\sqrt{X-3}}\right)^2=5^2\)=\(5^2\)

\(\Leftrightarrow-2\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=25-3=22\)

\(\Leftrightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}=11\)

Do \(\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\ge0\Rightarrow-\sqrt{\left(8-\sqrt{X-3}\right)\left(5-\sqrt{X-3}\right)}\le0\)

\(\Rightarrow\)PT vô nghiệm

Mình rút gọn như thế này đúng không nhỉ?

\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)

\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)

\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)

\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)

\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)

\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)

\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)

\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)

    \(\sqrt{x^2+8}-7x=\sqrt{x^2+3}-6\)(1)

\(\Leftrightarrow\sqrt{x^2+8}-3=7x-7+\sqrt{x^2+3}-2\)

\(\Leftrightarrow\frac{\left(\sqrt{x^2+8}-3\right)\left(\sqrt{x^2+8}+3\right)}{\left(\sqrt{x^2+8}+3\right)}=7\left(x-1\right)+\frac{\left(\sqrt{x^2+3}-2\right)\left(\sqrt{x^2+3}+2\right)}{\sqrt{x^2+3}+2}\)

\(\Leftrightarrow\frac{x^2+8-9}{\left(\sqrt{x^2+8}+3\right)}=7\left(x-1\right)+\frac{x^2-1}{\sqrt{x^2+3}+2}\)

\(\Leftrightarrow\frac{x^2-1}{\sqrt{x^2+8}+3}-7\left(x-1\right)-\frac{x^2-1}{\sqrt{x^2+3+2}}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt{x^2+8}+3}-7-\frac{x+1}{\sqrt{x^2+3}+2}\right)=0\)

\(\Leftrightarrow x-1=0\)

hay \(\frac{x+1}{\sqrt{x^2+8}+3}-7-\frac{x+1}{\sqrt{x^2+3}+2}=0\)(2)

Từ (1), có:

\(\sqrt{x^2+8}-\sqrt{x^2+3}=7x-6>0\)

\(\Leftrightarrow7x-6>0\)

\(\Leftrightarrow x>\frac{6}{7}\)

Khi đó, có:

     \(\frac{x+1}{\sqrt{x^2+8}+3}-\frac{\sqrt{x+1}}{\sqrt{x^2+3}+2}<0\)

\(\Rightarrow\frac{x+1}{\sqrt{x^2+8}+3}-\frac{x+1}{\sqrt{x^2+3}+2}-7<0\)

Vậy, pt (2) vô nghiệm

Do đó, pt (1) có 1 nghiệm là x = 1