\(\sqrt{x^2-3x+2}=\sqrt{10x-20}-\sqrt{x-3}\). MN giúp mình vs
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Mn ơi giải giúp mình pt này vs :
\(19+3x+4\sqrt{-x^2-x-6}=6\sqrt{2-x}+12\sqrt{2-x}+12\sqrt{x+3}\)
\(\sqrt{3x^2+6x+7}\) +\(\sqrt{5x^2+10x+14}\)=4-2x-\(x^2\)
2x^2+3x+\(\sqrt{2x^2+3x+9}\)=33
MN giúp e vs
Ta có \(\sqrt{3x^2+6x+7}=\sqrt{3\left(x+1\right)^2+4}\ge\sqrt{4}=2\)
Dấu"=" xảy ra khi x=-1
Tương tự \(\sqrt{5x^2+10x+14}=\sqrt{5\left(x+1\right)^2+9}\ge\sqrt{9}=3\)
Dấu"=" xảy ra khi x=-1
\(\Rightarrow4-2x-x^2\ge5\)
\(\Rightarrow-\left(x+1\right)^2+5\ge5\)
\(\Rightarrow\left(x+1\right)^2\le0\)
mà \(\left(x+1\right)^2\ge0\)
\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x=-1\)(tm)
Vậy....................
GPT \(\sqrt{x^2-3x+2}=\sqrt{10x-20}-\sqrt{x-3}\)
a)\(x=2\left(\sqrt{x-1}-\sqrt{x-2}\right)\)
b)\(\sqrt{5x^3+3x^2+3x-2}=\frac{1}{2}x^2+3x-\frac{1}{2}\)
mn ơi giúp mik vs ạ <3 !
a , \(x=2\left(\sqrt{x-1}-\sqrt{x-2}\right)\)
suy ra x =2
b, x=3
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\(3x^2+10x+\sqrt{3x+3}=x^3+26+\sqrt{5-2x}\)
giúp em với ạ
ĐKXĐ: \(-1\le x\le\dfrac{5}{2}\)
\(\Leftrightarrow\sqrt{3x+3}-3+1-\sqrt{5-2x}=x^3-3x^2-10x+24\)
\(\Leftrightarrow\dfrac{3\left(x-2\right)}{\sqrt{3x+3}+3}+\dfrac{2\left(x-2\right)}{1+\sqrt{5-2x}}=\left(x-2\right)\left(x-4\right)\left(x+3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{3}{\sqrt{3x+3}+3}+\dfrac{2}{1+\sqrt{5-2x}}=\left(x-4\right)\left(x+3\right)\left(1\right)\end{matrix}\right.\)
Xét (1), ta có:
\(\dfrac{3}{\sqrt{3x+3}+3}+\dfrac{2}{1+\sqrt{5-2x}}>0\)
\(-1\le x\le\dfrac{5}{2}\Rightarrow\left\{{}\begin{matrix}x+3>0\\x-4< 0\end{matrix}\right.\) \(\Rightarrow\left(x+3\right)\left(x-4\right)< 0\)
\(\Rightarrow\left(1\right)\) vô nghiệm hay pt có nghiệm duy nhất \(x=2\)
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1.A=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}\) và B=\(\dfrac{2\sqrt{x}}{\sqrt{x}-3}\) \(-\dfrac{3x+9}{x-9}\) với x ≥ 0,x ≠9
a) Tính giá trị biểu thức A khi x=16
b) Chứng minh A+3=\(\dfrac{3}{\sqrt{x}+3}\)
Mn giúp mk vs nhé ạ!!!
Giải phương trình:
a.\(\left(17-6x\right)\sqrt{3x-5}+\left(6x-7\right)\sqrt{7-3x}=2+8\sqrt{36x-9x^2-35}\)
b.\(\sqrt{x^2-3x+2}=\sqrt{10x-20}-\sqrt{x-3}\)
Mn giúp mình vs1, x^3-6x^2+10x-402, x^3+2x^2+2sqrt{2}x+2sqrt{2}03, x^4+x^2-sqrt{2}x+20
4, x^4+5x^3-12x^2+5x+105, left(x+5right)left(2x+12right)left(2x+20right)left(x+12right)3x^26, left(x^2-5x+1right)left(x^2-4right)6left(x-1right)^27, x^4-9x^3+16x^2+18x+40
Đọc tiếp
Mn giúp mình vs
1, \(x^3-6x^2+10x-4=0\)
2, \(x^3+2x^2+2\sqrt{2}x+2\sqrt{2}=0\)
3, \(x^4+x^2-\sqrt{2}x+2=0
\)
4, \(x^4+5x^3-12x^2+5x+1=0\)
5, \(\left(x+5\right)\left(2x+12\right)\left(2x+20\right)\left(x+12\right)=3x^2\)
6, \(\left(x^2-5x+1\right)\left(x^2-4\right)=6\left(x-1\right)^2\)
7, \(x^4-9x^3+16x^2+18x+4=0\)
1. \(x^3-6x^2+10x-4=0\)
<=> \(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
<=> \(\left(x-2\right)\left(x^2-4x+2\right)=0\)
<=> \(\orbr{\begin{cases}x=2\\x^2-4x+2=0\left(1\right)\end{cases}}\)
Giải pt (1): \(\Delta=\left(-4\right)^2-4.2=8>0\)
=> pt (1) có 2 nghiệm: \(x_1=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}\)
\(x_2=\frac{4-\sqrt{8}}{2}=2-\sqrt{2}\)
1) Ta có: \(x^3-6x^2+10x-4=0\)
\(\Leftrightarrow\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)-4x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-4x+2\right)\left(x-2\right)=0\)
+ \(x-2=0\)\(\Leftrightarrow\)\(x=2\)\(\left(TM\right)\)
+ \(x^2-4x+2=0\)\(\Leftrightarrow\)\(\left(x^2-4x+4\right)-2=0\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=2\)
\(\Leftrightarrow\)\(x-2=\pm\sqrt{2}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2+\sqrt{2}\approx3,4142\left(TM\right)\\x=2-\sqrt{2}\approx0,5858\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,5858;2;3,4142\right\}\)
4) \(x^4+5x^3-12x^2+5x+1=0\)
<=> \(\left(x^4-x^3\right)+\left(6x^3-6x^2\right)-\left(6x^2-6x\right)-\left(x-1\right)=0\)
<=> \(\left(x^3+6x^2-6x-1\right)\left(x-1\right)=0\)
<=> \(\left[\left(x-1\right)\left(x^2+x+1\right)-6x\left(x-1\right)\right]\left(x-1\right)=0\)
<=> \(\left(x-1\right)^2\left(x^2-5x+1\right)=0\)
<=> \(\orbr{\begin{cases}x=1\\x^2-5x+1=0\left(1\right)\end{cases}}\)
Giải pt (1) ta có: \(\Delta=\left(-5\right)^2-4=21>0\)
=> pt có 2 nghiệm
\(x_1=\frac{5+\sqrt{21}}{2}\); \(x_2=\frac{5-\sqrt{21}}{2}\)
Xem thêm câu trả lời
giải pt
\(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)
\(x^4+\sqrt{x^2+3}=3\)
\(\sqrt{x}+\sqrt[4]{20-x}=4\)
\(\sqrt{2x+1}+3\sqrt{4x^2-2x+1}=3+\sqrt{8x^3+1}\)
mn giải giúp e với nhé, e đang cần gấp. Thanks trc nha