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a: \(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{9}=\dfrac{11}{18}\)
hay \(x=\dfrac{11}{18}:\dfrac{1}{4}=\dfrac{11}{18}\cdot4=\dfrac{44}{18}=\dfrac{22}{9}\)
d: =>x+1;x-2 khác dấu
Trường hợp 1: \(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Leftrightarrow-1< x< 2\)
Trường hợp 2: \(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Leftrightarrow2< x< -1\left(loại\right)\)
e: =>x-2>0 hoặc x+2/3<0
=>x>2 hoặc x<-2/3
\(\Rightarrow\sqrt{y\left(2x-y\right)}.\sqrt{z\left(2y-z\right)}.\sqrt{x\left(2z-x\right)}=xyz\)
\(\Rightarrow\sqrt{xyz}.\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=xyz\)
\(\Rightarrow\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=\sqrt{xyz}\)
=>(2x-y)(2y-z)(2z-x)=xyz
=>(2x-y)2(2y-z)2(2z-x)2=x2y2z2
=>8(2x-y)2(2y-z)2(2z-x)2=8x2y2z2
(3-x2)(3-y2)(3-z2)
=3x2y2+3y2z2+3z2x2-x2y2z2
sau đó phân tích cái 8(2x-y)2(2y-z)2(2z-x)2
\(\Rightarrow\sqrt{y\left(2x-y\right)}.\sqrt{z\left(2y-z\right)}.\sqrt{x\left(2z-x\right)}=xyz\)
\(\Rightarrow\sqrt{xyz}.\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=xyz\)
\(\Rightarrow\sqrt{\left(2x-y\right)\left(2y-z\right)\left(2z-x\right)}=\sqrt{xyz}\)
=>(2x-y)(2y-z)(2z-x)=xyz
=>(2x-y)2(2y-z)2(2z-x)2=x2y2z2
=>8(2x-y)2(2y-z)2(2z-x)2=8x2y2z2
(3-x2)(3-y2)(3-z2)
=3x2y2+3y2z2+3z2x2-x2y2z2
sau đó phân tích cái 8(2x-y)2(2y-z)2(2z-x)2
a)
Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)
\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)
Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :
\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)
b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:
\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)
c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)
\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)
d) \((2x-1)^3=-27=(-3)^3\)
\(\Rightarrow 2x-1=-3\)
\(\Rightarrow 2x=-2\Rightarrow x=-1\)
e) \((x-2)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)
f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)
\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)
g) \((x-1)^2=(x-1)^6\)
\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)
\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)
\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x-1=-1\\ x-1=1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{0;1;2\right\}\)
a) \(\Leftrightarrow\left|x-3\right|=0;\left|y-2x\right|=0;\left|2z-x+y\right|=0\)
\(\Leftrightarrow x=3;y=2x;2z=-y+x\)
Ta có : y = 2x => y = 2 . 3 = 6
và 2z = -y + x => 2z = -6 + 3 = -3 => z = \(-\frac{3}{2}\)
b) \(\Leftrightarrow\left|x-y\right|+\left|2y+x-\frac{1}{2}\right|+\left|x+y+z\right|=0\) (vĩ mỗi số hạng trong tổng đều lớn hơn hoặc bằng 0)
\(\Leftrightarrow\left|x-y\right|=0;\left|2y+x-\frac{1}{2}\right|=0;\left|x+y+z\right|=0\)
\(\Leftrightarrow x=y;2y+x=\frac{1}{2};x+y=-z\)
Vì x = y nên \(2y+x=3y=\frac{1}{2}\Rightarrow x=y=\frac{1}{2}:3=\frac{1}{6}\)
và \(-z=x+y=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\Rightarrow z=-\frac{1}{3}\)
c) \(\left|2x-1\right|+\left|y+5\right|=0\)
Ta có:
\(\left\{{}\begin{matrix}\left|2x-1\right|\ge0\\\left|y+5\right|\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left|2x-1\right|+\left|y+5\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x-1\right|=0\\\left|y+5\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-1=0\\y+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=1\\y=0-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=-5\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{1}{2};-5\right\}.\)
Chúc bạn học tốt!
a) theo tính chất của dãy tỉ số bằng nhau có
\(\frac{x-y-z}{x}=\frac{-x+y-z}{y}=\frac{-x-y+z}{z}=\frac{x-y-z-x+y-z-x-y+z}{x+y+z}=\frac{-\left(x+y+z\right)}{x+y+z}=-1\)
=> x - y - z = - x => 2.x = y + z
y - x - z = - y => 2.y = x+z
z - x - y = - z => 2.z = x+y
Ta có: \(A=\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)=\frac{x+y}{x}.\frac{y+z}{y}.\frac{z+x}{z}=\frac{2z}{x}.\frac{2x}{y}.\frac{2y}{z}=\frac{2xyz}{xyz}=2\)
b) Vì \(\left|x+3y-1\right|\ge0\); \(-3\left|y+3\right|\le0\)
=> \(\left|x+3y-1\right|=-3\left|y+3\right|\) khi \(\left|x+3y-1\right|=-3\left|y+3\right|=0\)
=> x+ 3y - 1 = 0 và y + 3 = 0
=> x = 1 - 3y và y = -3 => x = 1- 3(-3) = 10; y = -3
=> C = 4.102.(-3) + 2.10.(-3)2 - (-3)2 = -1029
a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)
\(=-2x^4y^3+4x^3y^4-10x^2y^5\)
b) Ta có: \(\left(-2x\right)\cdot\left(x^3-3x^2-x+1\right)\)
\(=-2x^4+6x^3+2x^2-2x\)
c) Ta có: \(3x^2\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)
\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)
e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)
\(=-4x^3y^2+8x^2y^2-12x^2y\)
f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^3y\)
\(\left\{{}\begin{matrix}\left|x-2y-1\right|+5\ge5\\\dfrac{10}{\left|y-4\right|+2}\le5\end{matrix}\right.\)
Dấu "=" khi: \(\left\{{}\begin{matrix}y=4\\x=9\end{matrix}\right.\)
b) xem lại đề
a, \(\left|3x-4\right|+\left|3y+5\right|=0\)
Ta có :
\(\left|3x-4\right|\ge0\forall x;\left|3y+5\right|\ge0\forall x\\ \)
\(\Rightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\forall x\\ \Rightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=-\dfrac{5}{3}\end{matrix}\right.\\ Vậy.........\)
b, \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)
Ta có :
\(\left|x+\dfrac{19}{5}\right|\ge0\forall x;\left|y+\dfrac{1890}{1975}\right|\ge0\forall y;\left|z-2004\right|\ge0\forall z \)
\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1890}{1975}\\z=2004\end{matrix}\right.\\ Vậy............\)
c, \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)
Ta có : \(\left|x+\dfrac{9}{2}\right|\ge0\forall x;\left|y+\dfrac{4}{3}\right|\ge0\forall y;\left|z+\dfrac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\\ Vậy............\)
d, \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có :
\(\left|x+\dfrac{3}{4}\right|\ge0\forall x;\left|y-\dfrac{1}{5}\right|\ge0\forall y;\left|x+y+z\right|\ge0\forall x;y;z\)
\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x;y;z\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=0-\dfrac{1}{5}+\dfrac{3}{4}=\dfrac{11}{20}\end{matrix}\right.\\ Vậy.......\)
e, Câu cuối bn làm tương tự như câu a, b, c nhé!
a) \(3,6-\left|x-0,4\right|=0\)
\(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
Vậy \(x\in\left\{4;-3,2\right\}\)
b) Ta có:
\(\frac{x}{2}=y=\frac{z}{3}=\frac{2y}{2}=\frac{x-2y+z}{2-2+3}=\frac{210}{3}=70\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=70\\y=70\\\frac{z}{3}=70\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=140\\y=70\\z=210\end{matrix}\right.\)
Vậy \(x=140\); \(y=70\); \(z=210\)
c)\(\left|x+0,25\right|-4=\frac{1}{4}\)
\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=\frac{-17}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{-9}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{4;\frac{-9}{2}\right\}\)
d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,25\right)^4.\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,5\right)^8.\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,5\right)^{10}=\left(\frac{1}{2}\right)^{10}=\frac{1}{2^{10}}=\frac{1}{1024}\)
Vậy \(x=\frac{1}{1024}\)
e) \(3^{x-1}+5.3^{x-1}=162\)
\(\Leftrightarrow6.3^{x-1}=162\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
f) \(\frac{x}{-25}=\frac{2}{5}\)
\(\Leftrightarrow x=\left(-25\right).\frac{2}{5}=-10\)
Vậy \(x=-10\)
g) \(\left|x+\frac{3}{4}\right|-\frac{3}{4}=\sqrt{\frac{1}{9}}\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|-\frac{3}{4}=\frac{1}{3}\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{13}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{13}{12}\\x+\frac{3}{4}=-\frac{13}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-\frac{11}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};-\frac{11}{6}\right\}\)
a) \(3,6-\left|x-0,4\right|=0\)
\(\Rightarrow\left|x-0,4\right|=3,6-0\)
\(\Rightarrow\left|x-0,4\right|=3,6.\)
\(\Rightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,6+0,4\\x=\left(-3,6\right)+0,4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
Vậy \(x\in\left\{4;-3,2\right\}.\)
c) \(\left|x+0,25\right|-4=\frac{1}{4}\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{1}{4}+4\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=-\frac{17}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{17}{4}-\frac{1}{4}\\x=\left(-\frac{17}{4}\right)-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{9}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\frac{9}{2}\right\}.\)
d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)
\(\Rightarrow x:\left(0,25\right)^4=0,25\)
\(\Rightarrow x=\left(0,25\right).\left(0,25\right)^4\)
\(\Rightarrow x=\left(0,25\right)^5\)
\(\Rightarrow x=\frac{1}{1024}\)
Vậy \(x=\frac{1}{1024}.\)
Chúc bạn học tốt!
`@` `\text {Ans}`
`\downarrow`
Viết các biểu thức sau dưới dạng hiệu chứ ạ?
`e,`
`(x+1)(x-1)`
`= x(x-1) + x - 1`
`= x^2 - x + x - 1`
`= x^2 - 1`
`f,`
`(x-2y)(x+2y)?`
`= x(x+2y) - 2y(x+2y)`
`= x^2 + 2xy - 2xy - 4y^2`
`= x^2 - 4y^2`
`g,`
`(x+y+z)(x-y-z)`
`= x(x-y-z) + y(x-y-z) + z(x-y-z)`
`= x^2 - xy - xz + xy - y^2 - yz + xz - yz - z^2`
`= x^2 - y^2 - z^2 - 2yz`
`h,`
`(x-y+z)(x+y+z)`
`= x(x+y+z) - y(x+y+z) + z(x+y+z)`
`= x^2 + xy + xz - xy - y^2 - yz + xz + yz + z^2`
`= x^2 - y^2 + z^2 + 2xz`
Câu này c xem lại đề.