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*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)
\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)
\(M=6x^2+9xy-y^2-5x^2+2xy\)
\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)
\(M=x^2+11xy-y^2\)
* \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)
Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)
Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)
=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)
=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)
Thay x = 5/2 ; y = -4/3 vào M ta được :
\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)
\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)
\(M=\frac{-1159}{36}\)
Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3
Không chắc nha

a) \(5^{x+1}-2.5^x=375\)
\(\Rightarrow5^x\left(5-2\right)=375\)
\(\Rightarrow5^x.3=375\)
\(\Rightarrow5^x=125=5^3\)
\(\Rightarrow x=3\)
b) \(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow3^{2\left(x+1\right)}-5.3^{2x}=324\)
\(\Rightarrow3^2\left(3^{x+1}-5.3^x\right)=324\)
\(\Rightarrow9.3^x\left(3-5\right)=324\)
\(\Rightarrow3^x.\left(-2\right)=36\)
\(\Rightarrow3^x=-18=3^2.\left(-2\right)\)(vô lí vì 3x không chia hết cho 2)
c) \(\left(1-x\right)^5=32=2^5\)
\(\Rightarrow1-x=2\)
\(\Rightarrow x=-1\)
d) \(3.5^{2x+1}-3.25^x=300\)
\(\Rightarrow3\left(5^{2x}.5-5^{2x}\right)=300\)
\(\Rightarrow5^{2x}\left(5-1\right)=100\)
\(\Rightarrow5^{2x}.4=100\)
\(\Rightarrow5^{2x}=25=5^2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)

a, \(\frac{x}{4}=\frac{9}{10}\)
\(x.10=4.9\)
\(x.10=36\)
\(x=36:10=3,6\)
b, \(\frac{x}{24}=\frac{6}{x}\)
\(x.x=6.24\)
\(x^2=144=12^2\)
\(x=\pm12\)
c, \(\frac{5-2x}{4x-\frac{1}{-5}}\)
Thiếu đề.
d, \(\frac{10-2x}{6}=\frac{27}{5-x}\)
\(\frac{2\left(5-x\right)}{6}=\frac{27}{5-x}\)
\(\frac{5-x}{3}=\frac{27}{5-x}\)
\(\left(5-x\right)\left(5-x\right)=27.3\)
\(\left(5-x\right)^2=9^2\)
5 - x =9 hoặc 5 - x = -9
x = 5-9 hoặc x = 5+9
x= -4 hoặc x = 14
\(\frac{x}{4}=\frac{9}{10}\)
\(\Rightarrow10x=4\cdot9\)
\(\Rightarrow10x=36\)
\(\Rightarrow x=\frac{36}{10}=\frac{18}{5}\)
\(b,\frac{x}{24}=\frac{5}{x}\)
\(\Rightarrow x\cdot x=24\cdot5\)
\(\Rightarrow x^2=100\)
\(\Rightarrow x=\pm10\)

a/ \(\dfrac{x-1}{5}=\dfrac{1-2x}{3}\)
\(\Leftrightarrow3\left(x-1\right)=5\left(1-2x\right)\)
\(\Leftrightarrow3x-3=5-10x\)
\(\Leftrightarrow3x+10x=5+3\)
\(\Leftrightarrow13x=8\)
\(\Leftrightarrow x=\dfrac{8}{13}\)
Vậy ...
b/ \(\dfrac{3-\left|x\right|}{5}=1\dfrac{1}{2}:\dfrac{-6}{5}\)
\(\Leftrightarrow\dfrac{3-\left|x\right|}{5}=\dfrac{-5}{4}\)
\(\Leftrightarrow\left(3-\left|x\right|\right)4=5.\left(-5\right)\)
\(\Leftrightarrow\left(3-\left|x\right|\right).4=-25\)
\(\Leftrightarrow3-\left|x\right|=-6,25\)
\(\Leftrightarrow\left|x\right|=-3,25\)
\(\Leftrightarrow x\in\varnothing\)
\(\dfrac{x-1}{5}=\dfrac{1-2x}{3}\Rightarrow3x-3=5-10x\)
Áp dụng tính chất chuyển quế đổi giấu
3x+10x=5+3=8
13x=8
\(\Rightarrow\dfrac{8}{13}\)
b)\(\dfrac{3-|x|}{5}=1\dfrac{1}{2}chia\dfrac{-6}{5}=\dfrac{-5}{4}\)
3-/x/=5chia\(\dfrac{-5}{4}\)=-4
/x/=-4+3=-1
Mà /x/\(\ge0\Rightarrow x\in\varnothing\)
Tick em nha

\(A=\dfrac{x+5}{x-1}=\dfrac{x-1+6}{x-1}=\dfrac{x-1}{x-1}+\dfrac{6}{x-1}=1+\dfrac{6}{x-1}\)
\(\Rightarrow6⋮x-1\)
\(B=\dfrac{2x+4}{x+3}=\dfrac{2x+6-2}{x+3}=\dfrac{2x+6}{x+3}-\dfrac{2}{x+3}=2-\dfrac{2}{x+3}\)\(\Rightarrow2⋮x+3\)
\(C=\dfrac{6x+5}{2x-1}=\dfrac{6x-3+8}{2x-1}=\dfrac{6x-3}{2x-1}+\dfrac{8}{2x-1}=3+\dfrac{8}{2x-1}\)
\(\Rightarrow8⋮2x-1\)

Ukm, mk xin lỗi, bạn hiểu giùm mk nhá là f(5)=g(-3) đó, mk đánh nhầm. Cảm ơn bn, mong bn giúp mk!!!

a) \(\left|x+\frac{1}{5}\right|-4=-2\)
=) \(\left|x+\frac{1}{5}\right|=-2+4=2\)
=) \(x+\frac{1}{5}=2\)hoặc \(x+\frac{1}{5}=-2\)
=) \(x=2-\frac{1}{5}=\frac{9}{5}\); =) \(x=\left(-2\right)-\frac{1}{5}=\frac{-11}{5}\)
Vậy \(x=\left\{\frac{9}{5},\frac{-11}{5}\right\}\)
b)\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)
=) \(2x-\frac{6}{5}x=\frac{-1}{2}+\frac{1}{5}\)
=) \(x.\left(2-\frac{6}{5}\right)=\frac{-3}{10}\)
=) \(x.\frac{4}{5}=\frac{-3}{10}\)
=) \(x=\frac{-3}{10}:\frac{4}{5}\)
=) \(x=\frac{-3}{8}\)
c) \(\left(x-3\right)^{x+2}-\left(x-3\right)^{x+8}=0\)
=) \(\left(x-3\right)^{x+2}.\left(1-6\right)=0\)
=) \(\left(x-3\right)^{x+2}=0:\left(1-6\right)=0\)
Mà chỉ có \(0^x=0\)
=) \(x-3=0\)
=) \(x=0+3\)
=) \(x=3\)
a,
\(\left|x+\frac{1}{5}\right|-4=-2\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{5}=2\\x+\frac{1}{5}=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{9}{5}\\x=-\frac{11}{5}\end{cases}}\)
b,
\(2x-\frac{1}{5}=\frac{6}{5}x-\frac{1}{2}\)
\(\Rightarrow2x-\frac{6}{5}x=-\frac{1}{2}+\frac{1}{5}\)
\(\Rightarrow\frac{4}{5}x=-\frac{3}{10}\Leftrightarrow x=-\frac{3}{8}\)
c,
\(\left[x-3\right]^{x+2}-\left[x-3\right]^{x+8}=0\)
=> [x-3]x + 2 = [x-3]x+8
=> x + 2 = x + 8
=> x không tồn tại

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{2x+3y-4z}{2\cdot3+3\cdot4-4\cdot5}=\frac{-200}{-2}=100\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=100\\\frac{y}{4}=100\\\frac{z}{5}=100\end{cases}\Rightarrow\hept{\begin{cases}x=300\\y=400\\z=500\end{cases}}}\)
Vậy.......
Ta có: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{6}=\frac{3y}{12}=\frac{4z}{20}=\frac{2x+3y-4z}{6+12-20}=\frac{-200}{-2}=100\)
\(\Rightarrow x=100.3=300\)
\(y=100.4=400\)
\(z=100.5=500\)
Vậy x = 300; y = 400; z = 500
Lời giải:
\(|2x-5|+x=21\)
\(\Leftrightarrow |2x-5|=21-x\)
\(\Rightarrow \left[\begin{matrix} 2x-5=21-x\\ 2x-5=x-21\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{26}{3}\\ x=-16\end{matrix}\right.\) (đều thỏa mãn)
Vậy \(x=\frac{26}{3}; x=-16\)