A=1- (\(\text{ }\frac{\text{2x^2 - 1+x}}{\text{1-x^2}}\text{+}\text{ }\frac{\text{2x^3 - x +x^2}}{\text{1+x^2}}\)) * \(\frac{\text{(((1-x)(x^2-x)}}{\text{2x - 1}}\)
Rút gọn A và Cm A < 4/3
Những câu hỏi liên quan
Thực hiện phép tínha) frac{text{x + 9}}{x^2 - 9}-frac{text{3}}{text{x^2 + 3x}}b) frac{text{3x + 5
}}{text{x^2 - 5x
}}+frac{text{ 25 - x
}}{text{25 - 5x
}}c) frac{text{3
}}{text{2x
}}+frac{text{3x - 3
}}{text{2x - 1
}}+frac{ 2x^2 + 1
}{text{4x^2 - 2x
}}d) frac{text{1}}{text{3x - 2 }}-frac{1}{text{3x + 2 }}- frac{text{3x - 6}}{text{4 - 9x^2}}e) frac{text{18 }}{text{(x - 3)(x^2 - 9) }}-frac{text{3 }}{text{x^2 - 6x + 9 }}-frac{text{x}}{text{x^2 - 9}}g) frac{text{x + 2 }}{text{x + 3 }}-frac{text{...
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Thực hiện phép tính
a) \(\frac{\text{x + 9}}{x^2 - 9}-\frac{\text{3}}{\text{x^2 + 3x}}\)
b) \(\frac{\text{3x + 5 }}{\text{x^2 - 5x }}+\frac{\text{ 25 - x }}{\text{25 - 5x }}\)
c) \(\frac{\text{3 }}{\text{2x }}+\frac{\text{3x - 3 }}{\text{2x - 1 }}+\frac{ 2x^2 + 1 }{\text{4x^2 - 2x }}\)
d) \(\frac{\text{1}}{\text{3x - 2 }}-\frac{1}{\text{3x + 2 }}- \frac{\text{3x - 6}}{\text{4 - 9x^2}}\)
e) \(\frac{\text{18 }}{\text{(x - 3)(x^2 - 9) }}-\frac{\text{3 }}{\text{x^2 - 6x + 9 }}-\frac{\text{x}}{\text{x^2 - 9}}\)
g) \(\frac{\text{x + 2 }}{\text{x + 3 }}-\frac{\text{5 }}{\text{x^2 + x - 6 }}+\frac{\text{1}}{\text{2 - x}}\)
h) \(\frac{\text{4x }}{\text{x + 2 }}-\frac{\text{3x }}{\text{x - 2 }}+\frac{\text{12x}}{\text{x^2 - 4}}\)
i) \(\frac{\text{ x + 1 }}{\text{ x - 1 }}-\frac{\text{ x - 1 }}{\text{ x + 1 }}-\frac{\text{4}}{\text{1 - x^2}}\)
k) \(\frac{\text{
3x + 21
}}{\text{
x^2 - 9
}}+\frac{\text{2 }}{\text{x + 3 }}-\frac{\text{3}}{\text{x - 3}}\)
Ai giải giúp mấy bài toán vsBài 1:Asqrt{frac{1}{text{√}2+1}-frac{text{√}8-text{√}10}{2-text{√}5}}Bfrac{5text{√}5}{text{√}5+2}+frac{text{√}5}{text{√}5-1}-frac{3text{√}5}{3+text{√}5}Bài 2 rút gọn biểu thứcAleft(frac{x+sqrt[]{xy}}{text{√}x+text{√}y}-2right):frac{1}{text{√}x+2} với x :y 0Bleft(frac{a}{a-2text{√}a}+frac{a}{text{√}a-2}right):frac{text{√}a+1}{a-4text{√}a+4}Bài 3 cho biểu thứcPleft(frac{x-2}{x+2text{√}x}+frac{1}{text{√}x+2}right)frac{text{√}x+1}{text{√}x-1}a)Rút gọn Pb)tìm x để Ptext{√}...
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Ai giải giúp mấy bài toán vs
Bài 1:
A=\(\sqrt{\frac{1}{\text{√}2+1}-\frac{\text{√}8-\text{√}10}{2-\text{√}5}}\)
B=\(\frac{5\text{√}5}{\text{√}5+2}+\frac{\text{√}5}{\text{√}5-1}-\frac{3\text{√}5}{3+\text{√}5}\)
Bài 2 rút gọn biểu thức
A=\(\left(\frac{x+\sqrt[]{xy}}{\text{√}x+\text{√}y}-2\right):\frac{1}{\text{√}x+2}\) với x :y >0
B=\(\left(\frac{a}{a-2\text{√}a}+\frac{a}{\text{√}a-2}\right):\frac{\text{√}a+1}{a-4\text{√}a+4}\)
Bài 3 cho biểu thức
P=\(\left(\frac{x-2}{x+2\text{√}x}+\frac{1}{\text{√}x+2}\right)\frac{\text{√}x+1}{\text{√}x-1}\)
a)Rút gọn P
b)tìm x để P=\(\text{√}x+\frac{5}{2}\)
bài 4 rút gọn biểu thức
A=\(\frac{1}{x+\text{√}x}+\frac{2\text{√}x}{x-1}-\frac{1}{x-\text{√}x}\)
B=\(\left(\frac{x}{x+3\text{√}x}+\frac{1}{\text{√}x+3}\right):\left(1-\frac{2}{\text{√}x}+\frac{6}{x+3\text{√}x}\right)\)
Bài 5
A=\(\left(\frac{2}{\text{√}x-3}-\frac{1}{\text{√}x+3}-\frac{x}{\text{√}x\left(x-9\right)}\right):\text{(√}x+3-\frac{x}{\text{√}x-3}\)
a)rút gọn A
b)tìm gtri x để A= -1/4
AI GIẢI GIÙM MÌNH ĐI MÌNH TẠ ƠN
\(\left(\frac{2x\text{√}x+x-\text{√}x}{x\text{√}x-1}-\frac{x+\text{√}x}{x-1}\right)\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
a) Rút gọn
b) min
1.Giải pt sau:(sqrt{2} +2)(xsqrt{2} -1)2xsqrt{2} -sqrt{2}
2.Cho pt: 2(a-1).x-a(x-1)2a+3
3.Giải pt sau:
a) frac{2}{x+frac{text{1}}{text{1}+frac{x+text{1}}{x-2}}}frac{6}{3x-text{1}}
b) frac{frac{x+text{1}}{x-text{1}}-frac{x-text{1}}{x+text{1}}}{text{1}+frac{x+text{1}}{x-text{1}}}frac{x-text{1}}{2left(x+text{1}right)}
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1.Giải pt sau:(\(\sqrt{2}\) +2)(x\(\sqrt{2}\) -1)=2x\(\sqrt{2}\) -\(\sqrt{2}\)
2.Cho pt: 2(a-1).x-a(x-1)=2a+3
3.Giải pt sau:
a) \(\frac{2}{x+\frac{\text{1}}{\text{1}+\frac{x+\text{1}}{x-2}}}=\frac{6}{3x-\text{1}}\)
b) \(\frac{\frac{x+\text{1}}{x-\text{1}}-\frac{x-\text{1}}{x+\text{1}}}{\text{1}+\frac{x+\text{1}}{x-\text{1}}}=\frac{x-\text{1}}{2\left(x+\text{1}\right)}\)
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
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Mấy bài kia sao cái phương trình dài thê,s giải sao nổi
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22 tháng 11 2021 lúc 19:44
C=\(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{\text{5-x}}{\text{1-x}^{\text{2}}}\right)\):\(\dfrac{1-2x}{\text{x}^{\text{2}}-1}\)
a) Rút gọn
\(C=\dfrac{-\left(x+1\right)+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2}{1-2x}\)
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\(C=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Rightarrow C=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2\left(1-x\right)-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{-\left(1-x\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=-2.\dfrac{-1}{1-2x}\)
\(\Rightarrow C=\dfrac{2}{1-2x}\)
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\(P=\frac{2x+2}{\sqrt{x}}+\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}a;r\text{ú}tg\text{ọ}nPb;t\text{ì}mGTNNc\text{ủa}Pc;t\text{í}nhPt\text{ạ}ix=12+6\sqrt{3}\)
2 a. rút gọn biểu C = \(\dfrac{2x^{\text{2}}-x}{\text{x }-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)
b. Rút gọn biểu thức D = \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{\text{a}}-1}\right):\dfrac{\sqrt{\text{a}}+1}{a-2\sqrt{a}+1}\)
Vậy khi rút gọn một biểu thức hửu tỉ và một biểu thức chứa căn có tìm điều kiện xác định không?
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
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CHO BIỂU THỨC P=\(\text{[}\frac{X^2+1}{X^2-9}-\frac{X}{X+3}+\frac{5}{3-X}\text{]}\div\text{[}\frac{2X+10}{X+3}-1\text{]}\)1]
A ,RÚT GỌN P
B.TÍNH P KHI [X-1]=2
C.TÌM X ĐỂ P=\(\frac{X+5}{6}\)
ĐKXĐ: x \(\ne\)\(\pm\)3; x \(\ne\)-7
a) Ta có: P = \(\left(\frac{x^2+1}{x^2-9}-\frac{x}{x+3}+\frac{5}{3-x}\right):\left(\frac{2x+10}{x+3}-1\right)\)
P = \(\left(\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{2x+10-x-3}{x+3}\right)\)
P = \(\frac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}:\frac{x+7}{x+3}\)
P = \(\frac{-2x-14}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x+7}\)
P = \(\frac{-2\left(x+7\right)}{x-3}\cdot\frac{1}{x+7}=-\frac{2}{x-3}\)
b) Với x \(\ne\)\(\pm\)3 và x \(\ne\)-7
Ta có: x - 1 = 2 <=> x = 3 (ktm)
=> ko tồn tại giá trị P khi x - 1 = 2
c) Với x \(\ne\)\(\pm\)3; và x \(\ne\)-7
Ta có: P = \(\frac{x+5}{6}\)
<=> \(-\frac{2}{x-3}=\frac{x+5}{6}\)
=> (x - 3)(x + 5) = -12
<=> x2 + 2x - 15 = -12
<=> x2 + 2x - 3 = 0
<=> x2 + 3x - x - 3 = 0
<=> (x - 1)(x + 3) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\left(tm\right)\\x=-3\left(ktm\right)\end{cases}}\)
Vậy ...
a) \(P=\left(\frac{x^2+1}{x^2-9}-\frac{x}{x+3}+\frac{5}{3-x}\right):\left(\frac{2x+10}{x+3}-1\right)\left(x\ne\pm3\right)\)
\(=\left(\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\frac{x}{x+3}-\frac{5}{x-3}\right):\frac{2x+10-x-3}{x+3}\)
\(=\left(\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}-\frac{5x+15}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+7}{x+3}\)
\(=\frac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x+7}\)
\(=\frac{\left(-2x-14\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+7\right)}\)
\(=\frac{-2\left(x+7\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+7\right)}=-\frac{2}{x-3}\)
vậy \(P=-\frac{2}{x-3}\left(x\ne\pm3\right)\)
b) ta có \(P=-\frac{2}{x-3}\left(x\ne\pm3\right)\)
có x-1=2
<=> x=3 (không thỏa mãn điều kiện)
vậy không có giá trị P để x-1=2
c) ta có: \(P=-\frac{2}{x-3}\left(x\ne\pm3\right)\)
P=\(\frac{x+5}{6}\)=> \(\frac{-2}{x-3}=\frac{x+5}{6}\)
\(\Leftrightarrow x^2+2x-15=-12\)
\(\Leftrightarrow x^2+2x-3=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}}\)
đối chiếu điều kiện ta thấy x=1 thỏa mãn điều kiện
vậy \(P=\frac{x+5}{6}\)đạt được khi x=1
ĐKXĐ : \(x\ne\pm3\)
a) \(P=\left(\frac{x^2+1}{x^2-9}-\frac{x}{x+3}+\frac{5}{3-x}\right):\left(\frac{2x+10}{x+3}-1\right)\)
\(P=\left(\frac{x^2+1}{\left(x+3\right)\left(x-3\right)}-\frac{x}{x+3}-\frac{5}{x-3}\right):\left(\frac{2x+10}{x+3}-\frac{x+3}{x+3}\right)\)
\(P=\left(\frac{x^2+1}{\left(x+3\right)\left(x-3\right)}-\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{5\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{2x+10-x-3}{x+3}\right)\)
\(P=\left(\frac{x^2+1-x\left(x-3\right)-5\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{x+7}{x+3}\right)\)
\(P=\left(\frac{x^2+1-x^2+3x-5x-15}{\left(x+3\right)\left(x-3\right)}\right):\left(\frac{x+7}{x+3}\right)\)
\(P=\frac{-2x-14}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+7}\)
\(P=\frac{-2\left(x+7\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+7}\)
\(P=-\frac{2}{x-3}\)
b) \(P=-\frac{2}{x-3}\)( ĐKXĐ : \(x\ne3\))
x - 1 = 2 => x = 3 ( không tmđk )
Vậy không có giá trị của P khi x - 1 = 2
c) \(P=\frac{x+5}{6}\Leftrightarrow\frac{-2}{x-3}=\frac{x+5}{6}\)
<=> -2.6 = ( x - 3 )( x + 5 )
<=> -12 = x2 + 2x - 15
<=> x2 + 2x - 15 + 12 = 0
<=> x2 + 2x - 3 = 0
<=> x2 - x + 3x - 3 = 0
<=> x( x - 1 ) + 3( x - 1 ) = 0
<=> ( x + 3 )( x - 1 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
Vậy x = { -3 ; 1 }
Xem thêm câu trả lời
Tìm x biếta,left(frac{4}{5}right)^{2x+7}text{}frac{625}{256}b,frac{7^{x+2}+7^{x+1}+7^x}{57}text{}frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}c,left(4x-3right)^4text{}left(4x-3right)^2d,frac{2x+3}{5x+2}text{}frac{4x+5}{10x+2}e,frac{3x-1}{40-5x}text{}frac{2x-3x}{5x-34}f,frac{15}{x-9}text{}frac{20}{y-12}text{}frac{40}{z-2x} và xytext{}1200
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Tìm x biết
a,\(\left(\frac{4}{5}\right)^{2x+7}\text{=}\frac{625}{256}\)
b,\(\frac{7^{x+2}+7^{x+1}+7^x}{57}\text{=}\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
c,\(\left(4x-3\right)^4\text{=}\left(4x-3\right)^2\)
d,\(\frac{2x+3}{5x+2}\text{=}\frac{4x+5}{10x+2}\)
e,\(\frac{3x-1}{40-5x}\text{=}\frac{2x-3x}{5x-34}\)
f,\(\frac{15}{x-9}\text{=}\frac{20}{y-12}\text{=}\frac{40}{z-2x}\) và \(xy\text{=}1200\)
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
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