GPT: \(\sqrt{x-2\sqrt{x-3}-2}\)=1
1) GPT : \(\sqrt{x+2+2\sqrt{\text{x}+1}}+\sqrt{x+2-2\sqrt{x+1}}=\frac{x+5}{2}\)
2) GPT : \(\sqrt{x+2\sqrt{ }x-1}-\sqrt{x-2\sqrt{x-1}}=2\)
1/ ĐKXĐ:...
\(\Leftrightarrow\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-2\sqrt{x+1}+1}=\frac{x+5}{2}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(1-\sqrt{x+1}\right)^2}=\frac{x+5}{2}\)
\(\Leftrightarrow\sqrt{x+1}+1+\left|1-\sqrt{x+1}\right|=\frac{x+5}{2}\)
Nếu \(0\ge x\ge-1\Rightarrow\left|1-\sqrt{x+1}\right|=1-\sqrt{x+1}\)
\(\Rightarrow2=\frac{x+5}{2}\Leftrightarrow x=-1\left(tm\right)\)
Nếu \(x>0\Rightarrow\left|1-\sqrt{x+1}\right|=\sqrt{x+1}-1\)
\(\Rightarrow2\sqrt{x+1}=\frac{x+5}{2}\Leftrightarrow16x+16=x^2+10x+25\)
\(\Leftrightarrow x^2-6x+9=0\Leftrightarrow x=3\left(tm\right)\)
Vậy...
Câu dưới tương tự
Gpt: \(\sqrt[8]{1-x}+\sqrt[8]{1+x}+\sqrt[8]{1-x^2}=3\)
Đặt \(\left\{{}\begin{matrix}\sqrt[8]{1-x}=a\ge0\\\sqrt[8]{1+x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+ab=3\\a^8+b^8=2\end{matrix}\right.\)
Ta có: \(a^8+7+b^8+7\ge8a+8b\)
\(a^8+b^8+6\ge8ab\)
\(\Rightarrow2\left(a^8+b^8\right)+20\ge8\left(ab+a+b\right)=24\)
\(\Rightarrow a^8+b^8\ge2\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=1\) hay \(x=0\)
GPT:
1, \(6x^2+10x-92+\sqrt{\left(x+70\right)\left(2x^2+4x+16\right)}=0\)
2,\(x+3+\sqrt{1-x^2}=3\sqrt{x+1}+\sqrt{1-x}\)
ĐKXĐ:...
a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)
\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)
Pt trở thành:
\(3a^2-2b^2+ab=0\)
\(\Leftrightarrow\left(a+b\right)\left(3a-2b\right)=0\)
\(\Leftrightarrow3a=2b\)
\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)
\(\Leftrightarrow...\)
b. ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)
Phương trình trở thành:
\(a^2+2+ab=3a+b\)
\(\Leftrightarrow a^2-3a+2+ab-b=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a+b=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x+1}+\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow...\)
Gpt: \(\sqrt{x+5}+\sqrt{3-x}-2\left(\sqrt{15-2x-x^2}+1\right)=0\)
\(ĐK:-5\le x\le3\)
Đặt \(\sqrt{x+5}+\sqrt{3-x}=t\ge0\Leftrightarrow t^2-8=2\sqrt{15-2x-x^2}\), PTTT:
\(t-t^2+8-2=0\\ \Leftrightarrow t^2-t-6=0\\ \Leftrightarrow t=3\left(t\ge0\right)\\ \Leftrightarrow2\sqrt{15-2x-x^2}=3^2-8=1\\ \Leftrightarrow60-8x-4x^2=1\\ \Leftrightarrow4x^2+8x-59=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+3\sqrt{7}}{2}\left(tm\right)\\x=\dfrac{-2-3\sqrt{7}}{2}\left(tm\right)\end{matrix}\right.\)
Vậy nghiệm pt là ...
GPT : \(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}\)
\(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}\)
\(\Rightarrow\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{\left(x+1\right)\left(x+2\right)}\)
\(\Rightarrow\sqrt[3]{x+1}-1-\sqrt[3]{x+1}.\sqrt[3]{x+2}+\sqrt[3]{x+2}=0\)
\(\Rightarrow\left(\sqrt[3]{x+1}-1\right)-\sqrt[3]{x+2}\left(\sqrt[3]{x+1}-1\right)=0\)
\(\Rightarrow\left(\sqrt[3]{x+1}-1\right)\left(1-\sqrt[3]{x+2}\right)=0\)
Th1 : \(\sqrt[3]{x+1}-1=0\Rightarrow\sqrt[3]{x+1}=1\)
\(\Rightarrow x+1=1\Rightarrow x=0\)
Th2 : \(\sqrt[3]{x+2}-1=0\Rightarrow\sqrt[3]{x+2}=1\)
\(\Rightarrow x+2=1\Rightarrow x=-1\)
Vậy \(x\in\left\{0;-1\right\}\)
GPT : \(\frac{1}{\sqrt{x+3}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x}}\)
ĐK: \(x\ge0\)
\(PT\Leftrightarrow\frac{\sqrt{x+3}-\sqrt{x+2}}{1}+\frac{\sqrt{x+2}-\sqrt{x+1}}{1}+\frac{\sqrt{x+1}-\sqrt{x}}{1}=1\)
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow x+3+x-2\sqrt{x^2+3x}=1\)\(\Leftrightarrow2x+2=2\sqrt{x^2+3x}\)
\(\Leftrightarrow x^2+2x+1=x^2+3x\)
\(\Leftrightarrow x=1\)
Vậy.........................
Gpt: \(1+\frac{2}{3}\sqrt{x-x^2}=\sqrt{x}+\sqrt{1-x}\)
Đặt a = √(1-x)
b = √x
=> a2 + b2 = 1 và 1 + 2ab/3 = a + b
Giải hệ này tìm được a,b thế vô tìm được x
GPT : \(\sqrt{x-2}-\sqrt{x+1}=\sqrt{2x-1}-\sqrt{x+3}\)
GPT
\(\sqrt{x^2-2x+3}-\sqrt{x^2-6x+11}=\sqrt{3-x}-\sqrt{x-1}\)
Lời giải:
ĐK: $1\leq x\leq 3$
PT \(\Leftrightarrow \frac{x^2-2x+3-(x^2-6x+11)}{\sqrt{x^2-2x+3}+\sqrt{x^2-6x+11}}=\frac{3-x-(x-1)}{\sqrt{3-x}+\sqrt{x-1}}\)
\(\Leftrightarrow \frac{4(x-2)}{\sqrt{x^2-2x+3}+\sqrt{x^2-6x+11}}+\frac{2(x-2)}{\sqrt{3-x}+\sqrt{x-1}}=0\)
\(\Leftrightarrow (x-2)\left[\frac{4}{\sqrt{x^2-2x+3}+\sqrt{x^2-6x+11}}+\frac{2}{\sqrt{3-x}+\sqrt{x-1}}\right]=0\)
Dễ thấy biểu thức trong ngoặc vuông lớn hơn $0$ nên $x-2=0$
$\Rightarrow x=2$ (t/m)
Vậy.......
Lời giải:
ĐK: $1\leq x\leq 3$
PT \(\Leftrightarrow \frac{x^2-2x+3-(x^2-6x+11)}{\sqrt{x^2-2x+3}+\sqrt{x^2-6x+11}}=\frac{3-x-(x-1)}{\sqrt{3-x}+\sqrt{x-1}}\)
\(\Leftrightarrow \frac{4(x-2)}{\sqrt{x^2-2x+3}+\sqrt{x^2-6x+11}}+\frac{2(x-2)}{\sqrt{3-x}+\sqrt{x-1}}=0\)
\(\Leftrightarrow (x-2)\left[\frac{4}{\sqrt{x^2-2x+3}+\sqrt{x^2-6x+11}}+\frac{2}{\sqrt{3-x}+\sqrt{x-1}}\right]=0\)
Dễ thấy biểu thức trong ngoặc vuông lớn hơn $0$ nên $x-2=0$
$\Rightarrow x=2$ (t/m)
Vậy.......
GPT
\(\sqrt[]{2x+3+\sqrt{x+2}}+\sqrt[]{2x+2-\sqrt{2+x}}=1+2\sqrt{x+2}\)