Câu 2. (1,5 điểm) Giải các phương trình sau:
a) $\dfrac{5}{3} \sqrt{9 x+18}+\dfrac{1}{2} \sqrt{4 x+8}-15=\sqrt{2+x}$.
b) $\sqrt{x^{2}-4 x+4}-6=2 x$.
Câu 2: Giải các phương trình sau:
a. \(\sqrt{4x-8}\) - \(\sqrt{x-2}\) - 4 + \(\dfrac{1}{3}\)\(\sqrt{9x-18}\)
b. \(\sqrt{x^2-6x+9}\) - \(\dfrac{\sqrt{6+\sqrt{3}}}{\sqrt{2}+1}\)=0
b: Ta có: \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)
\(\Leftrightarrow x^2-6x+9=3\)
\(\Leftrightarrow x^2-6x+6=0\)
\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot6=36-24=12\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=3+\sqrt{3}\end{matrix}\right.\)
Giải các phương trình sau:
a) \(\sqrt{x^2-4+4}=2-x\)
b) \(\sqrt{4x-8}-\dfrac{1}{5}\sqrt{25x-50}=3\sqrt{x-2}-1\)
c) \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
d) \(\dfrac{1}{2}\sqrt{x-2}-4\sqrt{\dfrac{4x-8}{9}}+\sqrt{9x-18}-5=0\)
e)\(\sqrt{49-28x+4x^2}-5=0\)
f) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
g) x2 - 4x - 2\(\sqrt{2x-5}+5=0\)
h)\(\sqrt{3x-2}=\sqrt{x+1}\)
i) x + y + z + 8 = \(2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)
k) \(\sqrt{x^2-3x}-\sqrt{x-3}=0\)
l)\(\sqrt{x^2-4}+\sqrt{x-2}=0\)
m) \(4\sqrt{x+1}=x^2-5x+14\)
n) \(\sqrt{x^2-6x+9}-\sqrt{4x^2+4x+1}=0\)
c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)
\(\Leftrightarrow2\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=4\)
hay x=5
e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)
\(\Leftrightarrow\left|2x-7\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=5\\2x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
a. ĐKXĐ: $x\in\mathbb{R}$
PT $\Leftrightarrow \sqrt{(x-2)^2}=2-x$
$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$
$\Leftrightarrow x\leq 2$
b. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 2\sqrt{x-2}-\sqrt{x-2}=3\sqrt{x-2}-1$
$\Leftrightarrow 1=2\sqrt{x-2}$
$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$
$\Leftrightarrow \frac{1}{4}=x-2$
$\Leftrightarrow x=\frac{9}{4}$ (tm)
c. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4$
$\Leftrightarrow \sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4$
$\Leftrightarrow 2\sqrt{x-1}=4$
$\Leftrightarrow \sqrt{x-1}=2$
$\Leftrightarrow x-1=4$
$\Leftrightarrow x=5$ (tm)
d. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4}{9}}\sqrt{x-2}+\sqrt{9}.\sqrt{x-2}-5=0$
$\Leftrightarrow \frac{1}{2}\sqrt{x-2}-\frac{8}{3}\sqrt{x-2}+3\sqrt{x-2}-5=0$
$\Leftrightarrow \frac{5}{6}\sqrt{x-2}-5=0$
$\Leftrightarrow \sqrt{x-2}=6$
$\Leftrightarrow x-2=36$
$\Leftrightarrow x=38$ (tm)
Giải các phương trình sau theo phương pháp đặt ẩn phụ:
a.{\(\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\)
\(\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\)
b.{\(4\sqrt{x+3}-9\sqrt{y+1}=2\)
\(5\sqrt{x+3}+3\sqrt{y+1}=31\)
a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
Giải phương trình :
a) \(\sqrt{2x^2-\sqrt{2}x+\dfrac{1}{4}}=\sqrt{2}x\)
b)\(\sqrt{4x+8}+\dfrac{1}{3}\sqrt{9x+18}=3\sqrt{\dfrac{x+2}{4}}+\sqrt{2}\)
b: Ta có: \(\sqrt{4x+8}+\dfrac{1}{3}\sqrt{9x+18}=3\sqrt{\dfrac{x+2}{4}}+\sqrt{2}\)
\(\Leftrightarrow2\sqrt{x+2}+\dfrac{1}{3}\cdot3\sqrt{x+2}-\dfrac{3}{2}\sqrt{x+2}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{x+2}\cdot\dfrac{3}{2}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{2\sqrt{2}}{3}\)
\(\Leftrightarrow x+2=\dfrac{8}{9}\)
hay \(x=-\dfrac{10}{9}\)
Giải các phương trình sau:
\(a,\dfrac{3}{2}\sqrt{4+12x}-\dfrac{5}{3}\sqrt{9+27x}-\dfrac{1}{4}\sqrt{16+48x}=1\)
\(b,\sqrt{x^2-x+\dfrac{1}{4}}=3\)
a, ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{3}{2}.2\sqrt{1+3x}-\dfrac{5}{3}.3\sqrt{1+3x}-\dfrac{1}{4}.4\sqrt{1+3x}=1\\ \Leftrightarrow3\sqrt{1+3x}-5\sqrt{1+3x}-\sqrt{1+3x}=1\\ \Leftrightarrow-3\sqrt{1+3x}=1\\ \Leftrightarrow\sqrt{1+3x}=-\dfrac{1}{3}\left(vô.lí\right)\)
b, \(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)
\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)
Vậy \(S=\varnothing\)
b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
giải các phương trình sau:
\(1,\sqrt{18x}-6\sqrt{\dfrac{2x}{9}}=3-\sqrt{\dfrac{x}{2}}\)
\(2,\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\sqrt{27x}=-4\)
3, \(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
\(4,\sqrt{16x+16}-\sqrt{9x+9}=1\)
\(5,\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
\(6,\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=\dfrac{-2}{3}\)
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
Giải phương trình
\(a.\dfrac{3}{4}\sqrt{4x}-\sqrt{4x}+5=\dfrac{1}{4}\sqrt{4x}\)
\(b.\sqrt{3-x}-\sqrt{27-9x}+1,25.\sqrt{48-16x}=6\)
\(c.\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2,5}=\dfrac{2}{7}\)
\(d.\sqrt{9x^2+12x+4}=4\)
d. \(\sqrt{9x^2+12x+4}=4\)
<=> \(\sqrt{\left(3x+2\right)^2}=4\)
<=> \(|3x+2|=4\)
<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
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) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
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