Tìm x biết \(x^2\left(x+2\right)+4\left(x+2\right)=0\)\(\text{T\text{ì}m}xbi\text{ết}\)
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Các bạn giúp mình bài toán sau
\(\left(x+2\right)^3\text{-}\left(x+1\right)\left(x^2\text{-}x+1\right)=10\)
\(\left(x\text{-}1\right)^3\text{-}\left(x\text{-}2\right)\left(x^2+x+4\right).3x\left(x+1\right)=0\)
\(\left(x\text{-}3\right)^2\text{-}\left(x\text{-}2\right)\left(x^2+2x+4\right)\text{-}9x\left(x\text{-}1\right)=0\)
đề bài là tìm x à bạn? đề có cho điều kiện ko vậy ạ? (ví dụ như x nguyên?)
\(\left(x-1\right)^3+\left(x^3-8\right).3x.\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left[\left(x-1\right)^2+\left(x^3-8\right).3x\right]=0\)
TH1: \(x-1=0\Leftrightarrow x=1\)
TH2: \(\left(x-1\right)^2+\left(x^3-8\right).3x=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x^3-8\right).3x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x^3-8=0\\3x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\\left\{{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;2\right\}\)
\(choP=\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{2}{\sqrt{x}+1}\right):\left(\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\right)a;R\text{ú}tg\text{ọ}nP....b;T\text{í}nhPkhiX=3-2\sqrt{2}c;t\text{ì}mX\text{đ}\text{ể}P=1\)
tìm x, biết:
a) \(\left(3\text{x}+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
b) \(\left(3\text{x}-5\right)\left(7-5\text{x}\right)-\left(5\text{x}+2\right)\left(2-3\text{x}\right)=4\)
Tìm x, biết:
a) \(x\left(x-1\right)-x^2+2\text{x}=5\)
b) \(8\left(x-2\right)-2\left(3\text{x}-4\right)=2\)
c) \(\left(3\text{x}+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
d) \(\left(3\text{x}-5\right)\left(7-5\text{x}\right)-\left(5\text{x}+2\right)\left(2-3\text{x}\right)=4\)
\(\left(\dfrac{x}{2}+3\right).\left(5-6\text{x}\right)+\left(12\text{x}-2\right).\left(\dfrac{x}{4}+3\right)=0\)
Tìm x
\(\left(\dfrac{x}{2}+3\right)\left(5-6x\right)+\left(12x-2\right)\left(\dfrac{x}{4}+3\right)=0\)
\(\dfrac{5x}{2}-3x^2+15-18x+3x^2+36x-\dfrac{x}{2}-6=0\)
\(\dfrac{5x}{2}-\dfrac{x}{2}+18x+9=0\)
\(20x+9=0\)
\(x=\dfrac{-9}{20}\)
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Bài 1: Tính
Asqrt{5-2text{√}6}+sqrt{5+2text{√}6}
B left(sqrt{10}+sqrt{6}right)sqrt{8-2text{√}15}
Csqrt{4+text{√}7}+sqrt{4-text{√}7}
Dleft(3+text{√}5right)left(text{√}10-text{√}2right)sqrt{3-text{√}5}
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a0
b, x-2sqrt{xy}+y left(xge0;yge0right)
c, sqrt{xy}+2text{√}x-3text{√}y-6left(xge0;yge0right)
Bài 3: Rút gọn
M left(frac{1}{text{√}x-1}-frac{1}{text{√}x}right)divleft(frac{text{√}x+1}{text{√}x-2}-frac{text{√}x+2}{text{√}x-1}right)
a, Rút gọn...
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Bài 1: Tính
A=\(\sqrt{5-2\text{√}6}+\sqrt{5+2\text{√}6}\)
B= \(\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\text{√}15}\)
C=\(\sqrt{4+\text{√}7}+\sqrt{4-\text{√}7}\)
D=\(\left(3+\text{√}5\right)\left(\text{√}10-\text{√}2\right)\sqrt{3-\text{√}5}\)
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a>=0
b, x-2\(\sqrt{xy}\)+y \(\left(x\ge0;y\ge0\right)\)
c, \(\sqrt{xy}+2\text{√}x-3\text{√}y-6\)\(\left(x\ge0;y\ge0\right)\)
Bài 3: Rút gọn
M= \(\left(\frac{1}{\text{√}x-1}-\frac{1}{\text{√}x}\right)\div\left(\frac{\text{√}x+1}{\text{√}x-2}-\frac{\text{√}x+2}{\text{√}x-1}\right)\)
a, Rút gọn M
b, Tính giá trị của M khi x=2
c, Tìm x để M>0
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
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Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
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Bài 3:
a) ĐKXĐ:\(x>0; x\neq 1; x\neq 4\)
\(M=\frac{\sqrt{x}-(\sqrt{x}-1)}{(\sqrt{x}-1)\sqrt{x}}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{(x-1)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b)
Khi $x=2$ \(M=\frac{\sqrt{2}-2}{3\sqrt{2}}=\frac{1-\sqrt{2}}{3}\)
c)
Để \(M>0\leftrightarrow \frac{\sqrt{x}-2}{3\sqrt{x}}>0\leftrightarrow \sqrt{x}-2>0\leftrightarrow x>4\)
Kết hợp với ĐKXĐ suy ra $x>4$
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Làm tính chia:
a, \(\left(2^4+x^3-5\text{x}^2-3\text{x}-3\right)⋮\left(x^2-3\right)\)
b, \(\left(125x^3+8y^3\right):\left(25\text{x}^2-10\text{x}y+4y^2\right)\)
Tìm m nguyên để \(\dfrac{4\left(m^2+4\right)x-2mx+1}{x^2+2mx+4}\le\text{0}\text{ }\)vô nghiệm
Em kiểm tra lại tử số của đề bài
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text{Tìm }x,text{ }text{biết:}1text{)}text{ }2.left(x-frac{1}{3}right)-3left(x-frac{1}{2}right)frac{1}{2}x2text{) }-3left(x-frac{1}{4}right)-frac{1}{3}left(x+frac{1}{2}right)xtext{3) }frac{3}{2}left(x-frac{5}{3}right)-frac{4}{5}x+1text{4) }frac{1}{6}left(2.x-3right)frac{1}{2}left(-x+frac{1}{4}-frac{2}{3}right)text{5) }-frac{2}{3}left(x-frac{1}{4}right)frac{1}{3}left(2.x-1right)
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\(\text{Tìm }x,\text{ }\text{biết:}\)
\(1\text{)}\text{ }2.\left(x-\frac{1}{3}\right)-3\left(x-\frac{1}{2}\right)=\frac{1}{2}x\)
\(2\text{) }-3\left(x-\frac{1}{4}\right)-\frac{1}{3}\left(x+\frac{1}{2}\right)=x\)
\(\text{3) }\frac{3}{2}\left(x-\frac{5}{3}\right)-\frac{4}{5}=x+1\)
\(\text{4) }\frac{1}{6}\left(2.x-3\right)=\frac{1}{2}\left(-x+\frac{1}{4}-\frac{2}{3}\right)\)
\(\text{5) }-\frac{2}{3}\left(x-\frac{1}{4}\right)=\frac{1}{3}\left(2.x-1\right)\)