√36x-72 - 15 √x-2/25 = 4(5+√x-2)
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\(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\)
\(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=20+4\sqrt{x-2}\)
\(\Leftrightarrow6\sqrt{x-2}-3\sqrt{x-2}-4\sqrt{x-2}=20\)
\(\Leftrightarrow-\sqrt{x-2}=20\)(vô lý)
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\(\sqrt{36x-72}-15\sqrt{\frac{x-25}{25}}=4\left(5+\sqrt{x-2}\right)\)
\(\sqrt{36x-72}-15\sqrt{\frac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\)
Đặt \(\sqrt{x-2}=\:a\)(a >= 0)
Ta có 6a - 3a = 4(5 + a)
<=> a = - 20 (không thỏa điều kiện)
Vậy phương trình vô nghiệm
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bạn giải rõ hơn chút nữa được không? Mình cám ơn nhiều
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\(\sqrt{36-72x}\)- \(15\sqrt{\frac{x-2}{25}}\)= \(4\left(5+\sqrt{X+2}\right)\)
<=> \(\sqrt{36\left(x-2\right)}\)- \(15×\frac{\sqrt{x-2}}{5}\)= \(4\left(5+\sqrt{X+2}\right)\)
<=> \(6\sqrt{x-2}\)- \(3\sqrt{x-2}\) = \(4\left(5+\sqrt{X+2}\right)\)
<=> 6a - 3a = 20 + 4a
<=> a = -20
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Giải phương trình :
\(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\)
\(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\) \(\left(x\text{≥}2\right)\)
⇔ \(\sqrt{36\left(x-2\right)}-15.\dfrac{\sqrt{x-2}}{5}=20+4\sqrt{x-2}\)
⇔ \(6\sqrt{x-2}-3\sqrt{x-2}-4\sqrt{x-2}=20\)
⇔ \(-\sqrt{x-2}=20\) ( vô lý )
KL : Phương trình vô nghiệm .
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\(\sqrt{36x-72}-15\sqrt{\frac{X-2}{25}}=4\left(5+\sqrt{x-2}\right)\)Giải phương trình sau
giup mik voi can gap
Giai pt
a) 3✔x-5/2 - 2✔x-7/3 =✔x-1
b)✔36x-72 -15✔x-2/25 =4(5+✔x-2)
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b: \(\Leftrightarrow6\sqrt{x-2}-15\cdot\dfrac{\sqrt{x-2}}{5}=20+4\sqrt{x-2}\)
\(\Leftrightarrow\sqrt{x-2}\cdot6-3\sqrt{x-2}-4\sqrt{x-2}=20\)
\(\Leftrightarrow-\sqrt{x-2}=20\)(vô lý)
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tìm x biết
căn 36x-72+15căn x+2/25=4(5+căn x-2)
Giải PT:a) dfrac{1}{2}sqrt{x-1}-dfrac{3}{2}sqrt{9x-9}+24sqrt{dfrac{x-1}{64}}-17b) sqrt{18x-9}-0,5sqrt{2x-1}+dfrac{1}{2}sqrt{25left(2x-1right)}+sqrt{49left(2x-1right)}24c) sqrt{36x-72}-15sqrt{dfrac{x-2}{25}}4left(5+sqrt{x-2}right)d) sqrt{dfrac{1}{3x+2}}-dfrac{1}{2}sqrt{dfrac{9}{3x+2}}+sqrt{dfrac{16}{3x+2}}-5sqrt{dfrac{1}{12x+8}}1e) dfrac{1}{2}sqrt{dfrac{49x}{x+2}}-3sqrt{dfrac{x}{4x+8}}-sqrt{dfrac{x}{x+2}}-sqrt{5}0
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Giải PT:
a) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
b) \(\sqrt{18x-9}-0,5\sqrt{2x-1}+\dfrac{1}{2}\sqrt{25\left(2x-1\right)}+\sqrt{49\left(2x-1\right)}=24\)
c) \(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\)
d) \(\sqrt{\dfrac{1}{3x+2}}-\dfrac{1}{2}\sqrt{\dfrac{9}{3x+2}}+\sqrt{\dfrac{16}{3x+2}}-5\sqrt{\dfrac{1}{12x+8}}=1\)
e) \(\dfrac{1}{2}\sqrt{\dfrac{49x}{x+2}}-3\sqrt{\dfrac{x}{4x+8}}-\sqrt{\dfrac{x}{x+2}}-\sqrt{5}=0\)
a. ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$
$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$
$\Leftrightarrow -\sqrt{x-1}=-17$
$\Leftrightarrow \sqrt{x-1}=17$
$\Leftrightarrow x-1=289$
$\Leftrightarrow x=290$
b. ĐKXĐ: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$
$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$
$\Leftrihgtarrow \sqrt{2x-1}=2$
$\Leftrightarrow x=2,5$ (tm)
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c. ĐKXĐ: $x\geq 2$
PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm
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d. ĐKXĐ: $x>\frac{-2}{3}$
PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$
$\Leftrightarrow \frac{1}{3x+2}=1$
$\Leftrightarrow 3x+2=1$
$\Leftrightarrow x=-\frac{1}{3}$
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a) \(\sqrt{36x-72}-15\sqrt{\frac{x-2}{25}}=4\left(5+\sqrt{x-2}\right)\)
b) \(\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
c) \(\sqrt{3x^2}=x+2\)
Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{36(x-2)}-15\sqrt{\frac{1}{25}.(x-2)}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm.
b) ĐK: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{2x-2\sqrt{2x-1}}=2$
$\Leftrightarrow \sqrt{(2x-1)-2\sqrt{2x-1}+1}=2$
$\Leftrightarrow \sqrt{(\sqrt{2x-1}-1)^2}=2$
$\Leftrightarrow |\sqrt{2x-1}-1|=2$
$\Leftrightarrow \sqrt{2x-1}-1=\pm 2$
$\Leftrightarrow \sqrt{2x-1}=3$ (chọn) hoặc $\sqrt{2x-1}=-1$
$\Rightarrow x=5$ (thỏa mãn)
3.
PT \(\left\{\begin{matrix} x+2\geq 0\\ 3x^2=(x+2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ 2x^2-4x-4=0\end{matrix}\right.\Rightarrow x=1\pm \sqrt{3}\)
Lời giải:
a) ĐK: $x\geq 2$
PT $\Leftrightarrow \sqrt{36(x-2)}-15\sqrt{\frac{1}{25}.(x-2)}=4(5+\sqrt{x-2})$
$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$
$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)
Vậy pt vô nghiệm.
b) ĐK: $x\geq \frac{1}{2}$
PT $\Leftrightarrow \sqrt{2x-2\sqrt{2x-1}}=2$
$\Leftrightarrow \sqrt{(2x-1)-2\sqrt{2x-1}+1}=2$
$\Leftrightarrow \sqrt{(\sqrt{2x-1}-1)^2}=2$
$\Leftrightarrow |\sqrt{2x-1}-1|=2$
$\Leftrightarrow \sqrt{2x-1}-1=\pm 2$
$\Leftrightarrow \sqrt{2x-1}=3$ (chọn) hoặc $\sqrt{2x-1}=-1$
$\Rightarrow x=5$ (thỏa mãn)
3.
PT \(\left\{\begin{matrix} x+2\geq 0\\ 3x^2=(x+2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ 2x^2-4x-4=0\end{matrix}\right.\Rightarrow x=1\pm \sqrt{3}\)
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