\(a)\left(3\frac{1}{2}-x\right).1\frac{1}{4}=\frac{15}{6}\)

\(\left(\frac{7}{2}-x\right).\frac{5}{4}=\frac{15}{6}\)

\(\frac{7}{2}-x=\frac{15}{6}:\frac{5}{4}\)

\(\frac{7}{2}-x=2\)

\(x=\frac{7}{2}-2\)

\(\Rightarrow x=\frac{3}{2}\)

\(b)\frac{3}{2}-x=-\frac{7}{4}+5x\)

\(\frac{3}{2}-\frac{7}{4}=x+5x\)

\(-\frac{1}{4}=6x\)

\(x=-\frac{1}{4}:6\)

\(\Rightarrow x=-\frac{1}{24}\)

\(c)\frac{1}{2}x+150\%x=2016\)

\(\left(\frac{1}{2}+\frac{3}{2}\right).x=2016\)

\(2.x=2016\)

\(x=2016:2\)

\(\Rightarrow x=1008\)

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow2x^2+5x-1=7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}\)

\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)-7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a\\\sqrt{x-1}=b\end{matrix}\right.\)

\(\Rightarrow2a^2+3b^2-7ab=0\)

\(\Leftrightarrow\left(a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3b\\2a=b\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=9\left(x-1\right)\\4\left(x^2+x+1\right)=x-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

b/ ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}+2x-2\sqrt{x^2-4}-2=0\)

Đặt \(\sqrt{x-2}-\sqrt{x+2}=a< 0\)

\(\Rightarrow a^2=2x-2\sqrt{x^2-4}\)

Phương trình trở thành:

\(a+a^2-2=0\Leftrightarrow\left[{}\begin{matrix}a=1\left(l\right)\\a=-2\end{matrix}\right.\)

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

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

\(\Leftrightarrow x+2+4\sqrt{x-2}=x+2\)

\(\Leftrightarrow\sqrt{x-2}=0\)

e sửa lại câu 1 ạ: 2\(\text{x^2+5x−1=7√x3−1}\)

Bạn kiểm tra lại xem đã viết đúng đề chưa vậy?

Lời giải:

\(\lim_{x\to -1}\frac{\sqrt[3]{x}+x^2+x+1}{x+1}=\lim_{x\to -1}\frac{x(x+1)}{x+1}+\lim_{x\to -1}\frac{\sqrt[3]{x}+1}{x+1}\)

\(=\lim_{x\to -1}x+\lim_{x\to -1}\frac{x+1}{(x+1)(\sqrt[3]{x^2}-\sqrt[3]x+1)}\)

\(=\lim_{x\to -1}x+\lim_{x\to -1}\frac{1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}\)

\(=-1+\frac{1}{1-(-1)+1}=\frac{-2}{3}\)

a/

ĐKXĐ: \(x\ge\frac{1}{4}\)

\(\Leftrightarrow\sqrt{5x+1}\le\sqrt{4x-1}+3\sqrt{x}\)

\(\Leftrightarrow5x+1\le13x-1+6\sqrt{x\left(4x-1\right)}\)

\(\Leftrightarrow3\sqrt{x\left(4x-1\right)}\ge1-4x\)

Do \(x\ge\frac{1}{4}\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP\le0\end{matrix}\right.\) BPT luôn đúng

Vậy nghiệm của BPT đã cho là \(x\ge\frac{1}{4}\)

b/

ĐKXĐ: \(\left[{}\begin{matrix}x\ge\frac{-5+2\sqrt{5}}{5}\\x\le\frac{-5-2\sqrt{5}}{5}\end{matrix}\right.\)

Đặt \(\sqrt{5x^2+10x+1}=t\ge0\Rightarrow x^2+2x=\frac{t^2-1}{5}\)

BPT trở thành:

\(t\ge7-\frac{t^2-1}{5}\Leftrightarrow t^2+5t-36\ge0\)

\(\Rightarrow\left[{}\begin{matrix}t\le-9\left(l\right)\\t\ge4\end{matrix}\right.\)

\(\Rightarrow\sqrt{5x^2+10x+1}\ge4\)

\(\Leftrightarrow5x^2+10x-15\ge0\)

\(\Rightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)

c/

ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow x^2-4+1-\sqrt{x-1}+2-\sqrt{2x}< 0\)

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

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

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

\(\Leftrightarrow x-2< 0\Rightarrow x< 2\) (phần trong ngoặc to luôn dương)

Vậy nghiệm của BPT là \(1\le x< 2\)

+) \(\sqrt[3]{x+1}+\sqrt[3]{x-1}=\sqrt[3]{5x}\left(1\right)\)

+) Lập phương 2 vế ta được :

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

Thay ( 1 ) vào ( 2 ) ta có : 

\(\sqrt[3]{x^2-1}.\sqrt[3]{5x}=x\)

\(\Rightarrow4x^3-5x=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\x=\pm\frac{\sqrt{5}}{2}\end{cases}}\)

P/s : ko có tgian làm full . Thông cảm nhen ^-^

câu a : 

\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=1\dfrac{3}{4}\\ \left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{2}=\dfrac{7}{4}\\ \left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}+\dfrac{1}{2}\\ \left(x-\dfrac{1}{3}\right)^2=\dfrac{9}{4}\\ x-\dfrac{1}{3}=\sqrt{\dfrac{9}{4}}\\ x-\dfrac{1}{3}=\dfrac{3}{2}\\ x=\dfrac{3}{2}+\dfrac{1}{3}\\ x=\dfrac{11}{6}\)

câu b : 

\(\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\\ \Rightarrow\left(x-3\right)\cdot\left(x-3\right)=\left(-2\right)\cdot\left(-8\right)\\ \Rightarrow\left(x-3\right)^2=16\\ x-3=\sqrt{16}\\ x-3=4\\ x=4+3\\ x=7\)

câu c : 

\(\dfrac{9}{x}=\dfrac{-35}{105}\\ \Rightarrow\left(-35\right)\cdot x=9\cdot105\\ \left(-35\right)\cdot x=945\\ x=945\div\left(-35\right)\\ x=-27\)

`@` `\text {Ans}`

`\downarrow`

`1)`

\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)

`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(2x=\dfrac{7}{6}\)

`\Rightarrow`\(x=\dfrac{7}{6}\div2\)

`\Rightarrow`\(x=\dfrac{7}{12}\)

Vậy, `x = 7/12`

`2)`

\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)

`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)

`\Rightarrow`\(x=\dfrac{40}{21}\)

Vậy, `x = 40/21`

`3)`

\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)

`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)

`\Rightarrow`\(x=\dfrac{16}{21}\)

Vậy, `x = 16/21`

`4)`

\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{1}{12}\)

`\Rightarrow`\(x=\dfrac{1}{12}\div3\)

`\Rightarrow`\(x=\dfrac{1}{36}\)

Vậy, `x  = 1/36`

`5)`

\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)

`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)

`\Rightarrow`\(x=\dfrac{52}{21}\)

Vậy, `x = 52/21`

`6)`

\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)

`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(5x=\dfrac{1}{6}\)

`\Rightarrow`\(x=\dfrac{1}{6}\div5\)

`\Rightarrow`\(x=\dfrac{1}{30}\)

Vậy, `x = 1/30.`

\(\lim\limits_{x\rightarrow1}\dfrac{2-\sqrt[]{2x-1}\sqrt[3]{5x+3}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{2-2\sqrt[]{2x-1}+2\sqrt[]{2x-1}-\sqrt[]{2x-1}.\sqrt[3]{5x+3}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{2\left(1-\sqrt[]{2x-1}\right)+\sqrt[]{2x-1}\left(2-\sqrt[3]{5x+3}\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{-\dfrac{4\left(x-1\right)}{1+\sqrt[]{2x-1}}-\dfrac{5\sqrt[]{2x-1}\left(x-1\right)}{4+2\sqrt[3]{5x+3}+\sqrt[3]{\left(5x+3\right)^2}}}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\left(-\dfrac{4}{1+\sqrt[]{2x-1}}-\dfrac{5\sqrt[]{2x-1}}{4+2\sqrt[3]{5x+3}+\sqrt[3]{\left(5x+3\right)^2}}\right)\)

\(=-\dfrac{4}{1+1}-\dfrac{5\sqrt[]{1}}{4+4+4}=-\dfrac{29}{12}\)