tìm x biết:
(x-5)^2+(x+4)(3-x)=4
mình cần gấp
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\(\dfrac{5}{2}\times\dfrac{1}{3}+\dfrac{1}{4}\\ =\dfrac{5}{6}+\dfrac{1}{4}\\ =\dfrac{10}{12}+\dfrac{3}{12}\\ =\dfrac{13}{12}\)
= 5/6 + 1/4
= 10/12 + 3/12
= 13/12
Vậy giá trị biểu thức là 13/12
\(a,ĐK:x>0;x\ne1\\ b,A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ c,x=4\Leftrightarrow\sqrt{x}=2\Leftrightarrow A=\dfrac{2-1}{2}=\dfrac{1}{2}\)
tìm điều kiện xác định có thể rõ ràng chút được không ạ, chỗ này mình không hiểu lắm ý
a) \(x+215=480\)
\(x=480-215\)
\(x=265\)
b) \(725-x=185:5\)
\(725-x=37\)
\(x=725-37\)
\(x=688\)
c) \(x\times\dfrac{1}{3}+x\times\dfrac{4}{3}=\dfrac{2}{3}\)
\(x\times\left(\dfrac{1}{3}+\dfrac{4}{3}\right)=\dfrac{2}{3}\)
\(x\times\dfrac{5}{3}=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{3}\times\dfrac{3}{5}\)
\(x=\dfrac{2}{5}\)
d) \(\left(\dfrac{13}{5}+\dfrac{2}{3}\right):x=5\)
\(\dfrac{49}{15}:x=5\)
\(x=\dfrac{49}{15}:5=\dfrac{49}{15}\times\dfrac{1}{5}\)
\(x=\dfrac{49}{75}\)
a)x + 215 = 480
x = 480 - 215
x = 265
b) 725 - x = 185 : 5
725 - x = 37
x = 725 - 37
x = 688
Bài 2:
a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)
\(\Leftrightarrow x-2=4-x\)
\(\Leftrightarrow2x=6\)
hay x=3
b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)
\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)
\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)
a)\(\frac{x}{5}=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2\times5}{5}=2\)
Vậy .............
b) \(\frac{3}{8}=\frac{6}{x}\)
\(\Leftrightarrow x=\frac{8\times6}{3}=16\)
Vậy ................
c) \(\frac{1}{9}=\frac{x}{27}\)
\(\Leftrightarrow x=\frac{1\times27}{9}=3\)
Vậy ................
d) \(\frac{4}{x}=\frac{8}{6}\)
\(\Leftrightarrow x=\frac{4\times6}{8}=3\)
Vậy ............
e) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow3x+4x=20-6\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
Vậy .............
f) \(\frac{x}{-2}=\frac{-8}{x}\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
Vậy ...............
\(\frac{x}{5}=\frac{2}{5}\Rightarrow x=\frac{2\times5}{5}=2\)
\(\frac{1}{9}=\frac{x}{27}\Rightarrow x=\frac{1\times27}{9}=3\)
\(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{8\times6}{8}=6\)
\(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Rightarrow3.\left(x+2\right)=-4.\left(x-5\right)\)
\(\Rightarrow3x+6=-4x+20\)
\(\Rightarrow3x+4x=20-6\)
\(\Rightarrow7x=14\)
\(\Rightarrow x=2\)
a: \(\dfrac{x}{6}=\dfrac{8}{3}\)
=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)
b: \(\dfrac{5}{x}=\dfrac{4}{9}\)
=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)
c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)
=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)
=>x=-1-3=-4
d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)
=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)
=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)
=>\(x=-\dfrac{69}{8}\)
f: ĐKXĐ: x<>1
\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)
=>\(\left(x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)
a) 1/4(x-3)+2=1/5
1/4.(x-3) = 1/5-2
1/4.(x-3) = -9/5
x-3 = (-9/5):1/4
x-3 = -36/5
x = -36/5+3
x= -21/5
1: \(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z+7}{5}\)
mà x+y-z=8
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-1}{3}=\dfrac{y-2}{4}=\dfrac{z+7}{5}=\dfrac{x-1+y-2-z-7}{3+4-5}=\dfrac{8-3-7}{2}=\dfrac{-2}{2}=-1\)
=>\(\left\{{}\begin{matrix}x-1=-1\cdot3=-3\\y-2=-1\cdot4=-4\\z+7=-1\cdot5=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-2\\y=-2\\z=-12\end{matrix}\right.\)
2: \(\dfrac{x+1}{3}=\dfrac{y+2}{-4}=\dfrac{z-3}{5}\)
mà 3x+2y=47-42=5
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x+1}{3}=\dfrac{y+2}{-4}=\dfrac{z-3}{5}=\dfrac{3x+3+2y+4}{3\cdot3+2\left(-4\right)}=\dfrac{5+7}{9-8}=12\)
=>\(\left\{{}\begin{matrix}x+1=12\cdot3=36\\y+2=-12\cdot4=-48\\z-3=12\cdot5=60\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=35\\y=-48-2=-50\\z=60+3=63\end{matrix}\right.\)
Answer:
\(\left(x-5\right)^2+\left(x+4\right)\left(3-x\right)=4\)
\(\Rightarrow x^2-10x+25+3x-x^2+12-4x-4=0\)
\(\Rightarrow-11x+33=0\)
\(\Rightarrow x=3\)