a) \(\sqrt{x+8}=x+2\)
b) \(\sqrt{3+x}-2x=5+\sqrt{3+x}\)
c) \(\sqrt{5x+3}=3x-7\)
d) \(\sqrt{3x^2-2x-1}=3x+1\)
e) \(\sqrt{3x-5=4}\)
f) \(\sqrt{x^2-4}=x-1\)
g) \(\sqrt{x^2+x+1}=3-x\)
h) \(\sqrt{2x+2}=x-3\)
Giải phương trình:
a) \(\sqrt{x+2}=\sqrt{2x+1}+x\sqrt{x+2}\)
b) \(2+\sqrt{3-8x}=6x+\sqrt{4x-1}\)
c) \(\sqrt{10x+1}+\sqrt{3x-5}=\sqrt{9x+4}+\sqrt{2x-1}\)
d) \(1+\sqrt{x^2+4x}=\sqrt{x^2-3x+3}+\sqrt{2x^2+x+2}\)
e) \(\sqrt{x^2+15}=3x-2+\sqrt{x^2+8}\)
f) \(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\)
g) \(\sqrt{3x^2-7x+3}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)
h) \(\sqrt{2x^2+x-1}+\sqrt{3x^2+x-1}=\sqrt{x^2+4x-3}+\sqrt{2x^2+4x-3}\)
đa phần mình sử dụng phương pháp liên hợp nha bạn
\(\sqrt{a}-\sqrt{b}=\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
b. điều kiện \(\dfrac{1}{4}\le x\le\dfrac{3}{8}\), pt:
\(\Leftrightarrow\sqrt{3-8x}-\sqrt{4x-1}=6x-2\\ \Leftrightarrow\dfrac{3-8x-4x+1}{\sqrt{3-8x}+\sqrt{4x-1}}=2\left(3x-1\right)\\ \Leftrightarrow\dfrac{-4\left(3x-1\right)}{\sqrt{3-8x}+\sqrt{4x-1}}=2\left(3x-1\right)\\ \Leftrightarrow2\left(3x-1\right)+\dfrac{4\left(3x-1\right)}{\sqrt{3-8x}+\sqrt{4x-1}}=0\\ \Leftrightarrow2\left(3x-1\right)\left(1+\dfrac{2}{\sqrt{3-8x}+\sqrt{4x-1}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(n\right)\\1+\dfrac{2}{\sqrt{3-8x}+\sqrt{4x-1}}=0\left(vn\right)\end{matrix}\right.\)
d. điều kiện: \(x\le-4\cup x\ge0\), pt:
\(\Leftrightarrow1-\sqrt{x^2-3x+3}=\sqrt{2x^2+x+2}-\sqrt{x^2+4x}\\ \Leftrightarrow\dfrac{1-x^2+3x-3}{1+\sqrt{x^2-3x+3}}=\dfrac{2x^2+x+2-x^2-4x}{\sqrt{2x^2+x+2}+\sqrt{x^2+4x}}\\ \Leftrightarrow\dfrac{-\left(x-1\right)\left(x-2\right)}{1+\sqrt{x^2-3x+3}}=\dfrac{\left(x-1\right)\left(x-2\right)}{\sqrt{2x^2+x+2}+\sqrt{x^2+4x}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=1\left(n\right)\\\dfrac{-1}{1+\sqrt{x^2-3x+3}}=\dfrac{1}{\sqrt{2x^2+x+2}+\sqrt{x^2+4x}}\left(vn\right)\end{matrix}\right.\)
e. điều kiện:x thuộc R
\(\Leftrightarrow\sqrt{x^2+15}-4=3x-3+\sqrt{x^2+8}-3\\ \Leftrightarrow\dfrac{x^2+15-16}{\sqrt{x^2+15}+4}=3\left(x-1\right)+\dfrac{x^2+8-9}{\sqrt{x^2+8}+3}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+15}+4}-3\left(x-1\right)-\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+8}+3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\dfrac{\left(x+1\right)}{\sqrt{x^2+15}+4}-3-\dfrac{\left(x+1\right)}{\sqrt{x^2+8}+3}=0\left(1\right)\end{matrix}\right.\)
(1) mình không biết có vô nghiệm không nữa và cũng thua luôn
f. điều kiện: \(x\ge-2\)
bài này giải cách hơi khác một chút
đặt \(a=\sqrt{x+5}\left(\ge0\right)\\ b=\sqrt{x+2}\left(\ge0\right)\)
pt:
\(\Leftrightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left[\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)\right]\\ \Rightarrow\left(a-b\right)\left(1+ab\right)=3\left(1\right)\)
mà \(a^2-b^2=x+5-x-2=3\\ \Rightarrow\left(a-b\right)\left(a+b\right)=3\left(2\right)\)
=> (1) = (2)
\(\Leftrightarrow\left(a-b\right)\left(1+ab\right)=\left(a-b\right)\left(a+b\right)\\ \Leftrightarrow\left(a-b\right)\left(1+ab-a-b\right)=0\\ \Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\)
TH1: a=b \(\Leftrightarrow\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow x+5=x+2\left(vn\right)\)
TH2: a=1\(\Leftrightarrow\sqrt{x+5}=1\Leftrightarrow x=-4\left(l\right)\)
TH3: b=1\(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x=-1\left(n\right)\)
g. điều kiện: \(x\le-\sqrt{2}\cup x\ge\dfrac{7+\sqrt{37}}{2}\)
pt:
\(\dfrac{3x^2-7x+3-3x^2+5x+1}{\sqrt{3x^2-7x+2}+\sqrt{x^2-3x-4}}=\dfrac{x^2-2-x^2+3x-4}{\sqrt{3x^2-5x-1}+\sqrt{x^2-2}}\\ \Leftrightarrow\dfrac{-2\left(x-2\right)}{\sqrt{3x^2-7x+2}+\sqrt{x^2-3x-4}}=\dfrac{3\left(x-2\right)}{\sqrt{3x^2-5x-1}+\sqrt{x^2-2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(n\right)\\\dfrac{-2}{\sqrt{3x^2-7x+2}+\sqrt{x^2-3x-4}}=\dfrac{3}{\sqrt{3x^2-5x-1}+\sqrt{x^2-2}}\left(vn\right)\end{matrix}\right.\)h. điều kiện \(x\le-2-\sqrt{7}\cup x\ge-2+\sqrt{7}\)
\(\sqrt{2x^2+x-1}-\sqrt{x^2+4x-3}=\sqrt{2x^2+4x-3}-\sqrt{3x^2+x-1}\\ \Leftrightarrow\dfrac{2x^2+x-1-x^2-4x+3}{\sqrt{2x^2+x-1}+\sqrt{x^2+4x-3}}=\dfrac{2x^2+4x-3-3x^2-x+1}{\sqrt{2x^2+4x-3}+\sqrt{3x^2+x-1}}\\ \Leftrightarrow\dfrac{x^2-3x+2}{\sqrt{2x^2+x-1}+\sqrt{x^2+4x-3}}=\dfrac{-\left(x^2-3x+2\right)}{\sqrt{2x^2+4x-3}+\sqrt{3x^2+x-1}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=0\Leftrightarrow x=1\left(n\right),x=2\left(n\right)\\\dfrac{1}{\sqrt{2x^2+x-1}+\sqrt{x^2+4x-3}}=\dfrac{-1}{\sqrt{2x^2+4x-3}+\sqrt{3x^2+x-1}}\left(vn\right)\end{matrix}\right.\)
(nhớ tích cho mình nha, mấy bài kia mình ko biết làm huhu)
Giải các pt sau:
a) \(\sqrt{x+8}+\frac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)
b) \(x^4-2x^3+\sqrt{2x^3+x^2+2}-2=0\)
c) \(3x\sqrt[3]{x+7}\left(x+\sqrt[3]{x+7}\right)=7x^3+12x^2+5x-6\)
d) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
e) \(16x^2+19x+7+4\sqrt{-3x^2+5x+2}=\left(8x+2\right)\left(\sqrt{2-x}+2\sqrt{3x+1}\right)\)
f) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+8-\left(x+26\right)\sqrt{x-1}\)
g) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
a)\(\sqrt{1-x}\left(x-3x^2\right)=x^3-3x^2+2x+6\)
b)\(x^2+x+12\sqrt{x+1}=36\)
c)\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)
d)\(\sqrt{x^2+12}-3x=\sqrt{x^2+5}-5\)
e)\(4x^2+12+\sqrt{x-1}=4\left(x\sqrt{5x-1}+\sqrt{9-5x}\right)\)
f)\(4x^3-25x^2+43x+x\sqrt{3x-2}=22+\sqrt{3x-2}\)
g)\(2\left(x+1\right)\sqrt{x}+\sqrt{3\left(2x^3+5x^2+4x+1\right)}=5x^3-3x^2+8\)
h)\(\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)
i)\(\sqrt{1-3x}-\sqrt[3]{3x-1}=\left|6x-2\right|\)
k)\(\sqrt{2x^3+3x^2-1}=2x^2+2x-x^3-1\)
l)\(\sqrt{x^2+x-2}+x^2=\sqrt{2\left(x-1\right)}+1\)
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}\right)=0\)
\(\Rightarrow x=1\)
\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)
bạn làm nốt pần này nhá
1. Giải các phương trình sau:
a)\(\sqrt[4]{x-\sqrt{x^2-1}}+\sqrt[]{x+\sqrt{x^2-1}}=2\)
b)\(x^2-x-\sqrt{x^2-x+13}=7\)
c)\(x^2+2\sqrt{x^2-3x+1}=3x+4\)
d)\(2x^2+5\sqrt{x^2+3x+5}=23-6x\)
e)\(\sqrt{x^2+2x}=-2x^2-4x+3\)
f)\(\sqrt{\left(x+1\right)\left(x+2\right)}=x^2+3x+4\)
2. Giải các bất phương trình sau:
1)\(\sqrt{x^2-4x+5}\ge2x^2-8x\)
2)\(2x^2+4x+3\sqrt{3-2x-x^2}>1\)
3)\(\dfrac{\sqrt{-3x+16x-5}}{x-1}\le2\)
4)\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}\ge2\sqrt{x^2-5x+4}\)
5)\(\dfrac{9x^2-4}{\sqrt{5x^2-1}}\le3x+2\)
giai cac phuong trinh
a)\(2x^4+5x^3+x^2+5x+2=0\)
b)\(\sqrt{x-1}-\sqrt[3]{2-x}=1\)
c)\(x-\sqrt{x}+1=\sqrt{2x^2-30x+2}\)
d)\(2x^2+3x+7=\left(x-5\right)\sqrt{2x^2+1}\)
e)\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
giải pt
a) \(3\sqrt{x}+\frac{3}{2\sqrt{x}}=2x+\frac{1}{2x}-7\)
b) \(5\sqrt{x}+\frac{5}{2\sqrt{x}}=2x+\frac{1}{2x}+4\)
c) \(\sqrt{2x^2+8x+5}+\sqrt{2x^2-4x+5}=6\sqrt{x}\)
d) \(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\)
e) \(x^2+2x\sqrt{x-\frac{1}{x}}=3x+1\)
f) \(x^2-6x+x\sqrt{\frac{x^2-6}{x}}-6=0\)
g) \(\frac{3x^2}{3+\sqrt{x}}+6+2\sqrt{x}=5x\)
h) \(\frac{x^2}{4-3\sqrt{x}}+8=3\left(x+2\sqrt{x}\right)\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
e/ ĐKXĐ: ...
\(\Leftrightarrow x^2-1+2x\sqrt{\frac{x^2-1}{x}}=3x\)
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{x^2-1}{x}+2\sqrt{\frac{x^2-1}{x}}=3\)
Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)
\(a^2+2a=3\Leftrightarrow a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\frac{x^2-1}{x}}=1\Leftrightarrow x^2-1=x\Leftrightarrow x^2-x-1=0\)
f/ ĐKXĐ: ...
\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)
Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)
\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)
(Nghi binh 20/09)
Giải các phương trình sau:
a)\(32x^4-80x^3+50x^2+4x-3-4\sqrt{x-1}=0\)
b) \(\sqrt{5x^3-12x^2+12x-7}=\frac{x^2}{2}+2x-3\)
c)\(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=7\)
d)\(x+\sqrt{2x-3}=1+\sqrt{x-1}+\sqrt{x^2-3x+3}\)
e) \(\left(2x-1\right)\sqrt{x^2+1}=x^2+4x-5\)
f)\(\sqrt{8+\sqrt{x}}+\sqrt{5-\sqrt{x}}=5\)
g)\(2\left(x^2+2x+3\right)=5\sqrt{x^3+3x^2+3x+2}\)
h)\(\sqrt[3]{81x-8}=x^3-2x^2+\frac{4}{3}x-2\)
i)\(\sqrt{x\left(x+1\right)}+\sqrt{x\left(x+2\right)}=\sqrt{x\left(x-3\right)}\)
\(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=7\)
Ta đánh giá vế phải \(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=\sqrt{2\left(x-4\right)^2+9}+\sqrt{3\left(x-4\right)^2+16}\ge\sqrt{9}+\sqrt{16}=3+4=7\)(Do \(\left(x-4\right)^2\ge0\forall x\))
Như vậy, để \(\sqrt{2x^2-16x+41}+\sqrt{3x^2-24x+64}=7\)(hay dấu "=" xảy ra) thì \(\left(x-4\right)^2=0\)hay x = 4
Vậy nghiệm duy nhất của phương trình là 4
f, \(\sqrt{8+\sqrt{x}}+\sqrt{5-\sqrt{x}}=5\left(đk:25\ge x\ge0\right)\)
\(< =>\sqrt{8+\sqrt{x}}-\sqrt{9}+\sqrt{5-\sqrt{x}}-\sqrt{4}=0\)
\(< =>\frac{8+\sqrt{x}-9}{\sqrt{8+\sqrt{x}}+\sqrt{9}}+\frac{5-\sqrt{x}-4}{\sqrt{5-\sqrt{x}}+\sqrt{4}}=0\)
\(< =>\frac{\sqrt{x}-1}{\sqrt{8+\sqrt{x}}+\sqrt{9}}-\frac{\sqrt{x}-1}{\sqrt{5-\sqrt{x}}+\sqrt{4}}=0\)
\(< =>\left(\sqrt{x}-1\right)\left(\frac{1}{\sqrt{8+\sqrt{x}}+\sqrt{9}}-\frac{1}{\sqrt{5-\sqrt{x}}+\sqrt{4}}\right)=0\)
\(< =>x=1\)( dùng đk đánh giá cái ngoặc to nhé vì nó vô nghiệm )
giải phương trình
a, \(\sqrt{1+x}-\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)
b, \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
c, \(2x^2+4x=\sqrt{\dfrac{x+3}{2}}\)
d, \(2\left(x^2-3x+2\right)=3\sqrt{x^3+8}\)
e, \(729x^4+8\sqrt{1-x^2}=36\)
f, \(7x^2-10x+14=5\sqrt{x^4+4}\)
g, \(x^3+3x^2-3\sqrt[3]{3x+5}=1-3x\)
h, \(\sqrt{4-3\sqrt{10-3x}}=x-2\)
i, \(\sqrt{x-1}+\sqrt{x^2-1}=\sqrt{x^2-5x+4}\)
a) \(\sqrt{1+x}-\sqrt{8-x}+\sqrt{\left(1+x\right)\left(8-x\right)}=3\)
đặt t \(=\sqrt{1+x}-\sqrt{8-x}\)
\(\Leftrightarrow t^2=1+x-2\sqrt{\left(1+x\right)\left(8-x\right)}+8-x\)
\(\Leftrightarrow\sqrt{\left(1+x\right)\left(8-x\right)}=\dfrac{9-t^2}{2}\)
pt \(\Rightarrow t+\dfrac{9-t^2}{2}=3\)
\(\Leftrightarrow t^2-2t-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}-\sqrt{8-x}=-1\\\sqrt{1+x}-\sqrt{8+x}=3\end{matrix}\right.\)
suy ra tìm đc x
câu b đặt t =\(3x^2+5x+8\)
ta có pt \(\Leftrightarrow\sqrt{t}-\sqrt{t-7}=1\)
\(\Rightarrow t=16\)
\(\Leftrightarrow3x^2+5x+8=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{8}{3}\end{matrix}\right.\)
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp