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Bạn tự tìm ĐKXĐ.
a/ \(\sqrt{4-5x}=12\Rightarrow4-5x=144\Rightarrow x=-28\)
b/ \(10+\sqrt{3x}=\left(2+\sqrt{6}\right)^2=10+4\sqrt{6}\)
\(\Rightarrow\sqrt{3x}=4\sqrt{6}\Rightarrow\sqrt{x}=4\sqrt{2}\)
\(\Rightarrow x=32\)
c/ \(2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Rightarrow\sqrt{x+5}=2\Rightarrow x+5=4\Rightarrow x=-1\)
d/ \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)
e/ \(\sqrt{\frac{4x+3}{x+1}}=3\Leftrightarrow\frac{4x+3}{x+1}=9\)
\(\Rightarrow4x+3=9x+9\Rightarrow5x=-6\Rightarrow x=-\frac{6}{5}\)
f/ \(\sqrt{x-2}\le3\Rightarrow x-2\le9\Rightarrow2\le x\le11\)
a) \(\sqrt{1-4x+4x^2}=5\)
<=> \(\sqrt{4x^2-4x+1}=5\)
<=> 4x2 - 4x + 1 = 52
<=> 4x2 - 4x + 1 = 25
<=> 4x2 - 4x + 1 - 25 = 0
<=> 4x2 - 4x - 24 = 0
<=> 4(x + 2)(x - 3) = 0
<=> x = -2 hoặc x = 3
=> x = -2 hoặc x = 3
b) \(\sqrt{4-5x}=12\)
<=> \(\sqrt{-5x+4}=12\)
<=> -5x + 4 = 122
<=> -5x + 4 = 144
<=> -5x = 144 - 4
<=> -5x = 140
<=> x = -28
=> x = -28
\(a,\sqrt{1-4x+4x^2}=5\)
\(\Rightarrow4x^2-4x+1=25\)
\(\Rightarrow4x^2-4x-24=0\)
\(\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\)
\(\Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
\(b,\sqrt{4-5x}=12\)
\(\Rightarrow4-5x=144\)
\(\Rightarrow5x=-140\)
\(\Rightarrow x=-28\)
\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
<=> x + 1 = 16
<=> x = 15 (nhận)
~ ~ ~
\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow\sqrt{x+5}=2\)
<=> x + 5 = 4
<=> x = - 1 (nhận)
k) ĐK: $x^2\geq 5$
PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$
$\Leftrightarrow 2\sqrt{x^2-5}=4$
$\Leftrightarrow \sqrt{x^2-5}=2$
$\Rightarrow x^2-5=4$
$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)
l) ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$
$\Leftrightarrow 4\sqrt{x+1}=4$
$\Leftrightarrow \sqrt{x+1}=1$
$\Rightarrow x+1=1$
$\Rightarrow x=0$
m)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$
$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}=16$
$\Leftrightarrow \sqrt{x+1}=4$
$\Rightarrow x=15$ (thỏa mãn)
h)
ĐKXĐ: $x\geq -5$
PT $\Leftrightarrow \sqrt{x+5}=6$
$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)
i) ĐKXĐ: $x\geq 5$
PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)
\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)
j)
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$
$\Leftrightarrow -2\sqrt{2x}+4=0$
$\Leftrightarrow \sqrt{2x}=2$
$\Rightarrow x=2$ (thỏa mãn)
a, \(\sqrt{x-5} = 3 \)
<=> x - 5 = 9
<=> x = 14.
b, \(\sqrt{4-5x}=12\)
<=> 4 - 5x = 144
<=> 5x = -140
<=> x = -28.
c, \(\sqrt{x^{2}-6x+9}=3\)
<=> x2 - 6x + 9 = 9
<=> (x - 3)2 = 9
TH1:
x - 3 = 3
<=> x = 6.
TH2:
x - 3 = -3
<=> x = 0
d, \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3} \sqrt{9x+45}=4\)
<=> \(2\sqrt{x+5}+\sqrt{x+5}-\sqrt{x+5} = 4\)
<=> \(2\sqrt{x+5}\)= 4
<=> \(\sqrt{x+5}\) = 2
<=> x + 5 = 4
<=> x = -1.
\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)
\(\Leftrightarrow9x-7=7x+5\)
\(\Leftrightarrow9x-7x=5+7\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}\left(2+1-1\right)=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
\(\Leftrightarrow x=9\)
a) ĐK: \(x\ge5\)
\(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)
\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)
\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\) (t/m)
Vậy
b) \(-5x+7\sqrt{x}=-12\)
\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow\)\(\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
đến đây tự làm
c) d) e) bạn bình phương lên
f) \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)
\(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)
\(\ge\sqrt{9}+\sqrt{25}=8\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)
Vậy...
ĐKXĐ: bạn tự tìm
a/ Có vẻ bạn ghi nhầm đề, nhưng nói chung vẫn giải được, nghiệm xấu
\(\Leftrightarrow2\sqrt{x}+\frac{1}{2}\sqrt{x}-\frac{3}{4}\sqrt{5x}=5\)
\(\Leftrightarrow\sqrt{x}\left(\frac{5}{2}-\frac{3\sqrt{5}}{4}\right)=5\)
\(\Rightarrow\sqrt{x}=\frac{40+12\sqrt{5}}{11}\Rightarrow x=\left(\frac{40+12\sqrt{5}}{11}\right)^2\)
b/ \(\sqrt{3-x}-3\sqrt{3-x}+5\sqrt{3-x}=6\)
\(\Leftrightarrow3\sqrt{3-x}=6\)
\(\Leftrightarrow\sqrt{3-x}=2\Rightarrow3-x=4\Rightarrow x=-1\)
c/ \(7\left(5\sqrt{x}-2\right)=2\left(8\sqrt{x}+\frac{5}{2}\right)\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow19\sqrt{x}=19\)
\(\Rightarrow\sqrt{x}=1\Rightarrow x=1\)
d/ \(\sqrt{3x^2+12x+4}=4\)
\(\Leftrightarrow3x^2+12x+4=16\)
\(\Leftrightarrow3x^2+12x-12=0\)
\(\Rightarrow x=-2\pm2\sqrt{2}\)
\(dat:\sqrt{x-5}=a\Rightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\frac{1}{3}=\sqrt{9\left(x-5\right)}\Rightarrow\sqrt{4}.a+a-\frac{1}{3}=\sqrt{9}.a\Rightarrow3a-\frac{1}{3}=3a\left(voli\right)\Rightarrow vonghiem\)
câu a chắc đề như zầy pk bạn???
\(\sqrt{4x-20}+\sqrt{x-5}-\frac{1}{3}+\sqrt{9x-45}=4\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}+3\sqrt{x-5}=\frac{13}{3}\)
\(\Leftrightarrow6\sqrt{x-5}=\frac{13}{3}\Rightarrow\sqrt{x-5}=\frac{13}{18}\Leftrightarrow x=\frac{1789}{324}\)
b)đề như này đúng ko bạn??
\(\sqrt{16-32x}-\sqrt{12x}=\sqrt{3x}+\sqrt{9-18x}\)
\(\Leftrightarrow4\sqrt{1-2x}-2\sqrt{3x}=\sqrt{3x}+3\sqrt{1-2x}\)
\(\Leftrightarrow\sqrt{1-2x}-3\sqrt{3x}=0\Leftrightarrow\sqrt{1-2x}=3\sqrt{3x}\)
\(\Leftrightarrow1-2x=27x\Leftrightarrow x=\frac{1}{29}\)
câu c\(\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)
Xét điều kiện \(\left\{{}\begin{matrix}x\le1\\x\ge5\end{matrix}\right.\)không tồn tại số nào nằm trong khoảng này
Vậy pt trên vô nghiệm
a) \(\sqrt{4-5x}=12\)
ĐK : x ≤ 4/5
Bình phương hai vế
⇔ \(4-5x=144\)
⇔ \(-5x=140\)
⇔ \(x=-28\)( tm )
b) \(\sqrt{1-4x+4x^2}=5\)
⇔ \(\sqrt{\left(1-2x\right)^2}=5\)
⇔ \(\left|1-2x\right|=5\)
⇔ \(\orbr{\begin{cases}1-2x=5\\1-2x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)
ĐK : x ≥ -5
⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)
⇔ \(\left|2\right|\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot\left|3\right|\sqrt{x+5}=6\)
⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)
⇔ \(\frac{5}{4}\sqrt{x+5}=6\)
⇔ \(\sqrt{x+5}=\frac{24}{5}\)
⇔ \(x+5=\frac{576}{25}\)
⇔ \(x=\frac{451}{25}\)( tm )
d)\(\sqrt{x-2}\le3\)
ĐK : x ≥ 2
⇔ \(x-2\le9\)
⇔ \(x\le11\)
Kết hợp với điều kiện => Nghiệm của bpt là 2 ≤ x ≤ 11