ĐK: \(-\dfrac{1}{4}\le x\le3\)

\(pt\Leftrightarrow4x+1-6\sqrt{4x+1}+9+3-x-2\sqrt{3-x}+1=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{4x+1}=3\\\sqrt{3-x}=1\end{matrix}\right.\)

\(\Leftrightarrow x=2\left(tm\right)\)

a) ĐKXĐ : \(x\ge-3\)\(pt\Leftrightarrow x^2-2x+1=x+3-4\sqrt{x+3}+4\Leftrightarrow\left(x-1\right)^2=\left(\sqrt{x+3}-2\right)^2\Leftrightarrow x-1=\sqrt{x+3}-2\Leftrightarrow x+1=\sqrt{x+3}\Leftrightarrow\left(x+1\right)^2=x+3\left(x\ge-1\right)\Leftrightarrow x^2+2x+1=x+3\Leftrightarrow x^2+x-2=0\Leftrightarrow\left[{}\begin{matrix}x=1\left(tmdk\right)\\x=-2\left(kotm\right)\end{matrix}\right.\)

Bạn tự tìm điều kiện xác định nha :3

\(6\sqrt{4x+1}+2\sqrt{3-x}=3x+14\)

\(\Leftrightarrow-\left[4x+1-6\sqrt{4x+1}+9+3-x-2\sqrt{3-x}+1\right]+3x+14=3x+14\)\(\Leftrightarrow-\left[\left(\sqrt{4x+1}-3\right)^2+\left(\sqrt{3-x}-1\right)^2\right]=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{4x+1}-3=0\\\sqrt{3-x}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{4x+1}=3\\\sqrt{3-x}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x+1=9\\3-x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x=8\\x=2\end{matrix}\right.\Rightarrow x=2\)

Học tốt ạ

nhiều thế giải ko đổi đâu bạn

vậy trả lời câu a thôi

đkxđ : \(\frac{1}{2}\le x\le7\)

\(x^2-5x+3\sqrt{2x-1}=2\sqrt{14-2x}+5\)

\(\Leftrightarrow\left(x^2-5x\right)+3\left(\sqrt{2x-1}-3\right)=2\left(\sqrt{14-2x}-2\right)\)

\(\Leftrightarrow x\left(x-5\right)+\frac{3.\left(2x-10\right)}{\sqrt{2x-1}+3}+\frac{2.\left(2x-10\right)}{\sqrt{14-2x}+2}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+\frac{6}{\sqrt{2x-1}+3}+\frac{4}{\sqrt{14-2x}+2}\right)=0\)

\(\Leftrightarrow x=5\)

còn bài a,c lười đánh lắm

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)

\(ĐKXĐ:\frac{-1}{4}\le x\le3\)

\(PT\Leftrightarrow3x+14-6\sqrt{4x+1}-2\sqrt{3-x}=0\)

\(\Leftrightarrow\left(4x+1\right)-2.3\sqrt{4x+1}+9+\left(3-x\right)-3\sqrt{3-x}+1=0\)

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

Mà \(\left(\sqrt{4x+1}-3\right)^2\ge0\forall x\);\(\left(\sqrt{3-x}-1\right)^2\ge0\forall x\)

\(\Rightarrow\)(1) xảy ra khi \(\hept{\begin{cases}\sqrt{4x+1}=3\\\sqrt{3-x}=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x+1=9\\3-x=1\end{cases}}\Rightarrow x=2\left(tmđk\right)\)

Vậy nghiệm duy nhất của phương trình là 2

\(6\sqrt[]{4x+1}+2\sqrt[]{3-x}=3x+14\)

Ấn nhầm sorry Chờ làm lại

Câu a:

Ta có:

\((x-3)^2+x^4=-y^2+6y-4\)

\(\Leftrightarrow (x-3)^2+x^4+y^2-6y+4=0\)

\(\Leftrightarrow x^4+x^2-6x+9+y^2-6y+4=0\)

\(\Leftrightarrow x^4+x^2-6x+4+(y^2-6y+9)=0\)

\(\Leftrightarrow (x^4-2x^2+1)+3(x^2-2x+1)+(y^2-6y+9)=0\)

\(\Leftrightarrow (x^2-1)^2+3(x-1)^2+(y-3)^2=0\)

\(\Rightarrow (x^2-1)^2=(x-1)^2=(y-3)^2=0\)

\(\Rightarrow \left\{\begin{matrix} x=1\\ y=3\end{matrix}\right.\)

Vậy..........

Câu b:

ĐKXĐ: \(\frac{3}{2}\leq x\leq \frac{5}{2}\)

\(\sqrt{2x-3}+\sqrt{5-2x}-x^2+4x-6=0\)

\(\Leftrightarrow \sqrt{2x-3}+\sqrt{5-2x}=x^2-4x+6\)

Áp dụng BĐT Bunhiacopxky:

\(\text{VT}^2\leq (1+1)(2x-3+5-2x)=4\)

\(\Rightarrow \text{VT}\leq 2\)

Mà \(\text{VP}=x^2-4x+6=(x-2)^2+2\geq 2\)

Do đó để \(\text{VT}=\text{VP}\) thì \(\text{VT}=2=\text{VP}\)

Điều này xảy ra khi \(\left\{\begin{matrix} \sqrt{2x-3}=\sqrt{5-2x}\\ (x-2)^2=0\end{matrix}\right.\Rightarrow x=2\) (t/m)

Vậy pt có nghiệm duy nhất $x=2$

Câu c:

Áp dụng BĐT dạng \(|a|+|b|\geq |a+b|\) ta có:

\(\text{VP}\geq |(x-1)+(x-2)|+|2x-3|+|4x-14|=|2x-3|+|2x-3|+|2x-14|\)

\(=2(|2x-3|+|2x-7|)=2(|2x-3|+|7-2x|)\geq 2|2x-3+7-2x|=8\)

Và:

\(\text{VT}=4+4x-x^2=8-(x^2-4x+4)=8-(x-2)^2\leq 8\)

Do đó : \(\text{VT}\leq 8\leq \text{VP}\)

Dấu "=" xảy ra khi \((x-2)^2=0\Rightarrow x=2\)

Thử lại thấy đúng

Vậy PT có nghiệm duy nhất $x=2$

ĐK:  \(-\frac{1}{4}\le x\le3\)

Chuyển vế ta có

\(4x+1-2.\sqrt{4x+1}.3+9+3-x-2\sqrt{3-x}+1=0\)

\(\Leftrightarrow\left(\sqrt{4\text{ }x+1}-3\right)^2+\left(\sqrt{3-x}-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{4x+1}=3\\\sqrt{3-x}=1\end{cases}}\)

\(\hept{\begin{cases}4x+1=9\\3-x=1\end{cases}}\Leftrightarrow x=2\)(thỏa ĐK)

k giùm nha