ai làm hộ mk vs mk dang cần gấp
bài 1 tìm x
a)(x-3^2)^3=(3^3)^2 b3x/2=4y/5 va y-x=21 c)x-1/x+2=x-2/x+3
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ai làm hộ mk vs mk dang cần gấp
bài 1 tìm x
a)(x-3^2)^3=(3^3)^2 b3x/2=4y/5 va y-x=21 c)x-1/x+2=x-2/x+3
\(\left(x-\frac{3}{5}\right)=\frac{2}{5}×-\frac{1}{3}\)
\(\left(x-\frac{3}{5}\right)=-\frac{2}{165}\)
\(x=-\frac{2}{165}+\frac{3}{5}\)
\(x=\frac{97}{165}\)
vậy \(x=\frac{97}{165}\)
\(x×\left(\frac{3}{7}+\frac{2}{3}\right)=\frac{10}{21}\)
\(x×\frac{23}{21}=\frac{10}{21}\)
\(x=\frac{10}{21}:\frac{23}{21}\)
\(x=\frac{10}{23}\)
vậy \(x=\frac{10}{23}\)
\(\left(x-\frac{3}{5}\right):\frac{-1}{3}=\frac{2}{5}\)
=> \(x-\frac{3}{5}=\frac{2}{5}\cdot\left(-\frac{1}{3}\right)=-\frac{2}{15}\)
=> \(x=-\frac{2}{15}+\frac{3}{5}=-\frac{2}{15}+\frac{9}{15}=\frac{7}{15}\)
\(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\)
=> \(\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\)
=> \(-\frac{5}{21}x=\frac{10}{21}\)
=> \(x=\frac{10}{21}:\frac{-5}{21}=\frac{10}{21}\cdot\frac{-21}{5}=-2\)
Hai bài của ☆luffy cute☆ đều sai hết , xem xét lại đi nhé
a) \(x^2+y^2=\left(x+y\right)^2-2xy=1^2-2.\left(-6\right)=13\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=1^3-3.\left(-6\right).1=19\)
\(x^5+y^5=\left(x^2+y^2\right)\left(x^3+y^3\right)-x^2y^2\left(x+y\right)=13.19-\left(-6\right)^2.1=211\)
b) \(x^2+y^2=\left(x-y\right)^2+2xy=1^1+2.6=13\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)=1^3+3.6.1=19\)
\(x^5-y^5=\left(x^2+y^2\right)\left(x^3-y^3\right)+x^2y^2\left(x-y\right)=13.19+6^2.1=283\)
Bài 1:
Ta có:
\(y-x=25\Rightarrow y=25+x\)
Mà \(7x=4y\Rightarrow7x=4\cdot\left(25+x\right)\)
\(7x=100+4x\)
\(\Rightarrow7x-4x=100\)
\(3x=100\)
\(x=\frac{100}{3}\)
bài 1 :
Ta có: 7x=4y ⇔ x/4=y/7
áp dụng tính chất dãy tỉ số bằng nhau ta có
x/4=y/7=(y-x)/(7-4)=100/3
⇒x= 4 x 100/3=400/3 ; y = 7 x 100/3=700/3
bài 2
ta có x/5 = y/6 ⇔ x/20=y/24
y/8 = z/7 ⇔ y/24=z/21
⇒x/20=y/24=z/21
ADTCDTSBN(bài 1 có)
x/20=y/24=z/21=(x+y)/(20+24)=69/48=23/16
⇒x= 20 x 23/16 = 115/4
y= 24x 23/16=138/2
z=21x23/16=483/16
a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)
b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)
\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)
d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)
\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=3.\left(-1\right)=-3\\z=5.\left(-1\right)=-5\end{matrix}\right.\)
d: x+y=5
nên x=5-y
Ta có: xy=6
=>y(5-y)=6
=>y2-5y+6=0
=>(y-2)(y-3)=0
=>y=2 hoặc y=3
=>x=3 hoặc x=2
a: \(\Leftrightarrow\left(x-3;y+4\right)\in\left\{\left(1;-7\right);\left(-1;7\right);\left(-7;1\right);\left(7;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(4;-11\right);\left(2;3\right);\left(-4;-3\right);\left(10;-5\right)\right\}\)
\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà -3<x<30
nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)
b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)
mà -16<=x<20
nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)
c: \(\Leftrightarrow x-1+4⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)
d: \(\Leftrightarrow2x+4-5⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{-1;-3;3;-7\right\}\)
a) \(\left(x-3^2\right)^3=\left(3^3\right)^2\)
\(\left(x-3^2\right)^3=3^6\)
\(x-9=9\)
\(x=18\)
b) \(\frac{3x}{2}=\frac{4y}{5}\) và \(y-x=21\)
\(5\cdot3x=2\cdot4y\)
\(15x=8y\)
\(\frac{y}{15}=\frac{x}{8}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{y}{15}=\frac{x}{8}=\frac{y-x}{15-8}=\frac{21}{7}=3\)
\(\Rightarrow y=3\cdot15=45\)
\(\Rightarrow x=3\cdot8=24\)
c) \(x-\frac{1}{x}+2=x-\frac{2}{x}+3\)
\(x-\frac{1}{x}-x+\frac{2}{x}=3-2\)
\(\frac{1}{x}+\frac{2}{x}=1\)
\(\frac{3}{x}=1\)
\(x=3\)
không làm đâu mệt rùi