Tìm x: a, \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)
b, \(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=-2x\)
Những câu hỏi liên quan
giải các PT sau :a) left|2x+3right|-left|xright|+left|x-1right|2x+4b) sqrt{x}-dfrac{4}{sqrt{x+2}}+sqrt{x+2}0c) sqrt{x+sqrt{2x-1}}+sqrt{x-sqrt{2x-1}}sqrt{2}d) x+sqrt{x+dfrac{1}{2}+sqrt{x+dfrac{1}{4}}}4e) sqrt{4x+3}+sqrt{2x+1}6x+sqrt{8x^2+10x+3}-16f)sqrt[3]{x-2}+sqrt{x+1}3GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
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giải các PT sau :
a) \(\left|2x+3\right|-\left|x\right|+\left|x-1\right|=2x+4\)
b) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
c) \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
d) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=4\)
e) \(\sqrt{4x+3}+\sqrt{2x+1}=6x+\sqrt{8x^2+10x+3}-16\)
f)\(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP
giải pt , sqrt{x^4+4x^2}+sqrt{x+x^2}sqrt{left(x^2+sqrt{x}right)^2+9x^2}.x0x^30x^32.0.sqrt{0}x^32xsqrt{x}x^32xsqrt{x} 4left(x^3-2xsqrt{x}right)^204left(x^6-4x^4sqrt{x}+4x^2xright)04x^6-16x^4sqrt{x}+16x^2x04x^6+16x^316x^4sqrt{x}16x^4+4x^5+4x^6+16x^316x^4+4x^5+16x^4sqrt{x}4x^3left(x+1right)left(x^2+4right)4left(4x^4+4x^4sqrt{x}+x^4.xright)4x^3left(x+1right)left(x^2+4right)4left(2x^2+x^2sqrt{x}right)^22sqrt{2x^3left(x+1right)left(x^2+4right)}2left(2x^2+x^2sqrt{x}right)x^4+x^2+4x^2+x+2sqrt{2x^3le...
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giải pt , \(\sqrt{x^4+4x^2}+\sqrt{x+x^2}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}.\)
\(x=0\)
\(x^3=0\)
\(x^3=2.0.\sqrt{0}\)
\(x^3=2x\sqrt{x}\)
\(x^3=2x\sqrt{x}\)
\(4\left(x^3-2x\sqrt{x}\right)^2=0\)
\(4\left(x^6-4x^4\sqrt{x}+4x^2x\right)=0\)
\(4x^6-16x^4\sqrt{x}+16x^2x=0\)
\(4x^6+16x^3=16x^4\sqrt{x}\)
\(16x^4+4x^5+4x^6+16x^3=16x^4+4x^5+16x^4\sqrt{x}\)
\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(4x^4+4x^4\sqrt{x}+x^4.x\right)\)
\(4x^3\left(x+1\right)\left(x^2+4\right)=4\left(2x^2+x^2\sqrt{x}\right)^2\)
\(2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)\)
\(x^4+x^2+4x^2+x+2\sqrt{2x^3\left(x+1\right)\left(x^2+4\right)}=2\left(2x^2+x^2\sqrt{x}\right)+x^4+x^2+4x^2+x\)
\(\left(\sqrt{x^4+4x^2}+\sqrt{x^2+x}\right)^2=\left(x^4+2x^2\sqrt{x}+x\right)+9x^2\)
\(\sqrt{x^4+4x^2}+\sqrt{x^2+x}=\sqrt{\left(x^2+\sqrt{x}\right)^2+9x^2}\)
vậy x=0 là nghiệm của pt =))
cho mk hỏi một chút là đây đích thực có phải lớp 1 ko ak?
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25 tháng 4 2020 lúc 11:56
Gpt: a) sqrt[4]{3left(x+5right)}-sqrt[4]{11-x}sqrt[4]{13+x}-sqrt[4]{3left(3-xright)}
b) frac{1+2sqrt{x}-xsqrt{x}}{3-x-sqrt{2-x}}2left(frac{1+xsqrt{x}}{1+x}right) c) sqrt{x+1}+frac{4left(sqrt{x+1}+sqrt{x-2}right)}{3left(sqrt{x-2}+1right)^2}3
d) sqrt{frac{x-2}{x+1}}+frac{x+2}{left(sqrt{x+2}+sqrt{x-2}right)^2}1 e) 2x+1+xsqrt{x^2+2}+left(x+1right)sqrt{x^2+2x+2}0
f) sqrt{2x+3}cdotsqrt[3]{x+5}x^2+x-6
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Gpt: a) \(\sqrt[4]{3\left(x+5\right)}-\sqrt[4]{11-x}=\sqrt[4]{13+x}-\sqrt[4]{3\left(3-x\right)}\)
b) \(\frac{1+2\sqrt{x}-x\sqrt{x}}{3-x-\sqrt{2-x}}=2\left(\frac{1+x\sqrt{x}}{1+x}\right)\) c) \(\sqrt{x+1}+\frac{4\left(\sqrt{x+1}+\sqrt{x-2}\right)}{3\left(\sqrt{x-2}+1\right)^2}=3\)
d) \(\sqrt{\frac{x-2}{x+1}}+\frac{x+2}{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}=1\) e) \(2x+1+x\sqrt{x^2+2}+\left(x+1\right)\sqrt{x^2+2x+2}=0\)
f) \(\sqrt{2x+3}\cdot\sqrt[3]{x+5}=x^2+x-6\)
f) ĐKXĐ: \(x\ge-\frac{3}{2}\)
Khi đó VT > 0 nên \(VT>0\Rightarrow\left[{}\begin{matrix}x\ge2\\x\le-3\left(L\right)\end{matrix}\right.\)
Lũy thừa 6 cả 2 vế lên PT tương đương:
\( \left( x-3 \right) \left( {x}^{11}+9\,{x}^{10}+6\,{x}^{9}-142\,{x}^{ 8}-231\,{x}^{7}+1113\,{x}^{6}+2080\,{x}^{5}-4604\,{x}^{4}-6908\,{x}^{3 }+13222\,{x}^{2}+10983\,x-15327 \right) =0\)
Cái ngoặc to vô nghiệm vì nó tương đương:
\(\left( x-2 \right) ^{11}+31\, \left( x-2 \right) ^{10}+406\, \left( x -2 \right) ^{9}+2906\, \left( x-2 \right) ^{8}+12281\, \left( x-2 \right) ^{7}+31031\, \left( x-2 \right) ^{6}+46656\, \left( x-2 \right) ^{5}+46648\, \left( x-2 \right) ^{4}+46452\, \left( x-2 \right) ^{3}+44590\, \left( x-2 \right) ^{2}+36015\,x-55223 = 0\)(vô nghiệm với mọi \(x\ge2\))
Vậy x = 3.
PS: Nghiệm đẹp thế này chắc có cách AM-Gm độc đáo nhưng mình chưa nghĩ ra
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@Akai Haruma, @Nguyễn Việt Lâm
giúp em vs ạ! Cần gấp ạ
em cảm ơn nhiều!
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giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
Rút gọna.left(2sqrt{x}+sqrt{2x}right)left(sqrt{x}-sqrt{2x}right)b. left(sqrt{3x}+sqrt{2x}right)left(3sqrt{x}-sqrt{6x}right)c.left(frac{4}{3}sqrt{3}+sqrt{2}sqrt{3frac{1}{3}}right)left(sqrt{1,2}+sqrt{2}-4sqrt{frac{1}{3}}right)-2d.left(2sqrt{x}+sqrt{y}right)left(3sqrt{x}-2sqrt{y}right)(x,y lớn hơn hoặc bằng 0)e.left(sqrt{x}+sqrt{y}right)left(x-sqrt{x}sqrt{y}+sqrt{y}right) (x,y lớn hơn hoặc bằng 0)
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Rút gọn
a.\(\left(2\sqrt{x}+\sqrt{2x}\right)\left(\sqrt{x}-\sqrt{2x}\right)\)
b. \(\left(\sqrt{3x}+\sqrt{2x}\right)\left(3\sqrt{x}-\sqrt{6x}\right)\)
c.\(\left(\frac{4}{3}\sqrt{3}+\sqrt{2}\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{3}}\right)-2\)
d.\(\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)\)(x,y lớn hơn hoặc bằng 0)
e.\(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{x}\sqrt{y}+\sqrt{y}\right)\) (x,y lớn hơn hoặc bằng 0)
a:dfrac{b}{left(a-4right)^2}.sqrt{dfrac{left(a-4right)^4}{b^2}}left(b0;ane4right)
b:dfrac{xsqrt{x}-ysqrt{y}}{sqrt{x}-sqrt{y}}left(xge0;yge0;xne0right)
c:dfrac{a}{left(b-2right)^2}.sqrt{dfrac{left(b-2right)^4}{a^2}left(a0;bne2right)}
d:dfrac{x}{left(y-3right)^2}.sqrt{dfrac{left(y-3right)^2}{x^2}left(x0;yne3right)}
e:2x +dfrac{sqrt{1-6x+9x^2}}{3x-1}
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a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
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a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
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Tìm x:
a.\(\sqrt{4-\sqrt{4+x}}=x\)
b.\(4\left(\sqrt{x-1}-3\right)x^2+\left(13\sqrt{x+1}-8\right)x-4\sqrt{x-1}-3=0\)
c.\(\sqrt{2x-3}+2\sqrt{x-3}\ge3\sqrt[4]{2x^2+x-6}\)
Tìm x
a)\(\sqrt{x-1}=2\left(x\ge1\right)\)
b)\(\sqrt{3-x}=4\left(x\le3\right)\)
c)\(2.\sqrt{3-2x}=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\)
d)\(4-\sqrt{x-1}=\dfrac{1}{2}\left(x\ge1\right)\)
e)\(\sqrt{x-1}-3=1\)
f)\(\dfrac{1}{2}-2.\sqrt{x+2}=\dfrac{1}{4}\)
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
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a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
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Giải các phương trình sau:1) sqrt{3x^2+5x+8}-sqrt{3x^2+5x+1}12) x^2-2x-12+4sqrt{left(4-xright)left(2+xright)}03) 3sqrt{x}+dfrac{3}{2sqrt{x}}2x+dfrac{1}{2x}-74) sqrt{x}-dfrac{4}{sqrt{x+2}}+sqrt{x+2}05)left(x-7right)sqrt{dfrac{x+3}{x-7}}x+46) 2sqrt{x-4}+sqrt{x-1}sqrt{2x-3}+sqrt{4x-16}7) sqrt{x+2sqrt{x-1}}+sqrt{x-2sqrt{x-1}}dfrac{x+3}{2}Giúp mình với ajk, mink đang cần gấp
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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
23 tháng 11 2019 lúc 23:16
giải pt
a) \(x+\sqrt{4-x^2}-3x\sqrt{4-x^2}=2\)
b) \(2\left(\sqrt{4-x^2}+4\right)\left(\sqrt{x+2}+\sqrt{2-x}\right)-5=0\)
c) \(\left(\sqrt{x^2-4}-x+1\right)\left(\sqrt{x-2}+\sqrt{x+2}\right)+2=0\)
d) \(\sqrt{x+2}-\sqrt{x-1}=\frac{6}{\sqrt{x^2+x-2}-x}\)
e) \(\frac{2}{\sqrt{x-1}+\sqrt{3-x}}=1+\sqrt{3+2x-x^2}\)
a/ ĐKXĐ: \(-2\le x\le2\)
Đặt \(x+\sqrt{4-x^2}=a\Rightarrow a^2=4+2x\sqrt{4-x^2}\Rightarrow x\sqrt{4-x^2}=\frac{a^2-4}{2}\)
\(\Rightarrow a-\frac{3\left(a^2-4\right)}{2}=2\)
\(\Leftrightarrow-3a^2+2a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\sqrt{4-x^2}=2\\x+\sqrt{4-x^2}=-\frac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{4-x^2}=2-x\\3\sqrt{4-x^2}=-4-3x\left(x\le-\frac{4}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x^2=x^2-4x+4\\12\left(4-x^2\right)=9x^2+24x+16\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-4x=0\\21x^2+24x-32=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\frac{-12+4\sqrt{51}}{2}\left(l\right)\\x=\frac{-12-4\sqrt{51}}{2}\end{matrix}\right.\)
Mấy câu còn lại và bài kia tầm 30ph nữa sẽ làm, bận chút xíu việc
b/ ĐKXĐ: \(-2\le x\le2\)
\(\Leftrightarrow\left(2\sqrt{4-x^2}+4+4\right)\left(\sqrt{x+2}+\sqrt{2-x}\right)-5=0\)
Đặt \(\sqrt{x+2}+\sqrt{2-x}=a>0\Rightarrow a^2=4+2\sqrt{4-x^2}\)
\(\Rightarrow\left(a^2+4\right)a-5=0\)
\(\Leftrightarrow a^3+4a-5=0\Leftrightarrow\left(a-1\right)\left(a^2+a+5\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{x+2}+\sqrt{2-x}=1\)
\(\Leftrightarrow4+2\sqrt{4-x^2}=1\Rightarrow2\sqrt{4-x^2}=-3\)
Vậy pt vô nghiệm
Thật ra bài này có thể biện luận vô nghiệm ngay từ đầu:
\(\sqrt{x+2}+\sqrt{2-x}\ge\sqrt{x+2+2-x}=2\)
\(2\left(\sqrt{4-x^2}+4\right)\ge2.4=8\)
\(\Rightarrow VT>8.2-5=11>0\) nên pt vô nghiệm
c/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\left(\sqrt{x^2-4}-x+1\right)\left(\sqrt{x-2}+\sqrt{x+2}\right)=-2\)
Do \(\sqrt{x+2}>\sqrt{x-2}\Rightarrow\sqrt{x+2}-\sqrt{x-2}\ne0\)
Nhân cả 2 vế của pt với \(\sqrt{x+2}-\sqrt{x-2}\) và rút gọn ta được:
\(4\left(\sqrt{x^2-4}-x+1\right)=-2\left(\sqrt{x+2}-\sqrt{x-2}\right)\)
Đặt \(\sqrt{x+2}-\sqrt{x-2}=a>0\)
\(\Rightarrow a^2=2x-2\sqrt{x^2-4}\Rightarrow\sqrt{x^2-4}-x=-\frac{a^2}{2}\)
Phương trình trở thành:
\(4\left(-\frac{a^2}{2}+1\right)=-2a\)
\(\Leftrightarrow-a^2+2=-a\Leftrightarrow a^2-a-2=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+2}-\sqrt{x-2}=2\Leftrightarrow\sqrt{x+2}=2+\sqrt{x-2}\)
\(\Leftrightarrow x+2=x+2+4\sqrt{x-2}\)
\(\Rightarrow4\sqrt{x-2}=0\Rightarrow x=2\)
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