Tìm biết:
Đáp số: .
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\(x+\dfrac{4}{5}\times\dfrac{3}{8}=\dfrac{3}{2}\)
\(x+\dfrac{3}{10}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}-\dfrac{3}{10}\)
\(x=\dfrac{12}{10}=\dfrac{6}{5}\)
a: \(x\in\left\{25;30;35\right\}\)
b: \(x\in\left\{24;32;40;48;56;64\right\}\)
c: \(x\in\left\{3;4;6\right\}\)
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: =>xy=-18
=>x,y khác dấu
mà x<y<0
nên không có giá trị nào của x và y thỏa mãn yêu cầu đề bài
b: =>(x+1)(y-2)=3
\(\Leftrightarrow\left(x+1,y-2\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
hay \(\left(x,y\right)\in\left\{\left(0;5\right);\left(2;3\right);\left(-2;-1\right);\left(-4;1\right)\right\}\)
c: \(\Leftrightarrow8x-4=3x-9\)
=>5x=-5
hay x=-1
Đáp án D
x + 73 : 5 = 20 + 7 , 5 × 4 x + 73 : 5 = 20 + 30 x + 73 : 5 = 50 x + 73 = 250 x = 250 - 73 x = 177
Vậy x=177
Đáp án cần chọn là: D
\(a,\Rightarrow2\left(x+5\right)-12=-20\\ \Rightarrow2\left(x+5\right)=-8\\ \Rightarrow x+5=-4\Rightarrow x=-9\\ b,\Rightarrow x-28+x=-24\\ \Rightarrow2x=4\Rightarrow x=2\\ c,\Rightarrow6\left(x-2\right)^3=-384\\ \Rightarrow\left(x-2\right)^3=-64=\left(-4\right)^3\\ \Rightarrow x-2=-4\Rightarrow x=-2\\ d,\Rightarrow2^x\left(1+2^3\right)=72\\ \Rightarrow2^x\cdot9=72\\ \Rightarrow2^x=8=2^3\Rightarrow x=3\)
a,⇒2(x+5)−12=−20⇒2(x+5)=−8⇒x+5=−4⇒x=−9b,⇒x−28+x=−24⇒2x=4⇒x=2c,⇒6(x−2)3=−384⇒(x−2)3=−64=(−4)3⇒x−2=−4⇒x=−2d,⇒2x(1+23)=72⇒2x⋅9=72⇒2x=8=23⇒x=3a,⇒2(x+5)−12=−20⇒2(x+5)=−8⇒x+5=−4⇒x=−9b,⇒x−28+x=−24⇒2x=4⇒x=2c,⇒6(x−2)3=−384⇒(x−2)3=−64=(−4)3⇒x−2=−4⇒x=−2d,⇒2x(1+23)=72⇒2x⋅9=72⇒2x=8=23⇒x=3
\(\dfrac{25}{3}.x=\dfrac{5}{6}+\dfrac{4}{3}\)
\(\dfrac{25}{3}.x=\dfrac{13}{6}\)
\(x=\dfrac{13}{3}:\dfrac{25}{3}\)
\(x=\dfrac{39}{75}\)
\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)
a. (x + 2)2 + (x + 3)2 - 2(x - 2)(x - 3) = 19
<=> (x2 + 4x + 4) + (x2 + 6x + 9) - (2x + 4)(x - 3) = 19
<=> x2 + 4x + 4 + x2 + 6x + 9 - 2x2 + 6x - 4x + 12 = 19
<=> x2 + x2 - 2x2 + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0
<=> 12x + 6 = 0
<=> 6(2x + 1) = 0
<=> 2x + 1 = 0
<=> 2x = -1
<=> x = \(\dfrac{-1}{2}\)
\(\left(x+72\right):5=20+7,5\times4\)
\(\left(x+72\right):5=20+30\)
\(\left(x+72\right):5=50\)
\(x+72=50\times5\)
\(x+72=250\)
\(x=250-72\)
\(x=178\)