A, x*2+x*3=50
B, x*9-x*5=24
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PV
2
R
17 tháng 11 2021
Bài 5:
a)x+37=50
⇔x=13
b)2x-3=11
⇔2x=14
⇔x=7
c)(2+x):5=6
⇔2+x=30
⇔x=28
d)2+x:5=6
⇔x:5=4
⇔x=20
17 tháng 11 2021
a)x+37=50
x =50-37
x =13
b)2.x-3=11
2.x =11+3
2.x =14
x =14:2
x =7
c)(2+x):5=6
2+x =6.5
2+x =30
x=30-2
x=28
d)2+x:5=6
x:5=6-2
x:5=4
x=4.5
x=20
Câu trả lời nhớ ghi đầy đủ như này nha bạn
24 tháng 11 2021
\(a,x< 50\Leftrightarrow\sqrt{x}-1< 5\sqrt{2}-1\\ M=\dfrac{\sqrt{x}-1}{2}\in Z\\ \Leftrightarrow\sqrt{x}-1\in B\left(2\right)=\left\{0;2;4;6\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;5;7\right\}\\ \Leftrightarrow x\in\left\{1;9;25;49\right\}\\ b,\Leftrightarrow\sqrt{x}-5\inƯ\left(9\right)=\left\{-3;-1;1;3;9\right\}\left(\sqrt{x}-5>-5\right)\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;6;8;14\right\}\\ \Leftrightarrow x\in\left\{4;16;36;64;196\right\}\)
QP
1
13 tháng 12 2022
b: 30 chia hết cho x
45 chia hết cho x
Do đó: \(x\inƯC\left(30;45\right)=Ư\left(15\right)\)
mà x>10
nen x=15
c: \(\Leftrightarrow x+2\in\left\{1;-1;3;-3;9;-9\right\}\)
hay \(x\in\left\{-1;-3;1;-5;7;-11\right\}\)
d: =>x+3+14 chia hết cho x+3
=>\(x+3\in\left\{1;-1;2;-2;7;-7;14;-14\right\}\)
hay \(x\in\left\{-2;-4;-1;-5;4;-10;11;-17\right\}\)
14 tháng 10 2021
\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)
26 tháng 12 2021
\(a,\left(x-4\right).5=50\\ \Rightarrow x-4=10\\ \Rightarrow x=14\\ b,\left(-470-x\right)\left(-3\right)=-120\\ \Rightarrow-470-x=40\\ \Rightarrow x=-510\\ c,5x^3+\left(-170\right)=-210\\ \Rightarrow5x^3=-40\\ \Rightarrow x^3=-8\\ \Rightarrow x=-2\)
TT
26 tháng 12 2021
x - 4 = 50 : 5
x - 4 = 10
x = 10 + 4
x = 14
b) -470 - x = -120: (-3)
-470 - x = 40
x = -470 - 40
x = -510
c) 5. x3 = -210 - (-170)
5. x3 = -40
x3 = -40: 5
x3 = -8
x = -2
Cho mình tim nha
2 tháng 10 2021
a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)
\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)
\(\Leftrightarrow-62x=-92\)
hay \(x=\dfrac{46}{31}\)
b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)
\(\Leftrightarrow8x^2+49x-15=0\)
\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)
14 tháng 12 2022
a: x chia hết cho 4;5;10
nên \(x\in BC\left(4;5;10\right)\)
mà 10<=x<50
nên x=40
b: x=33
DQ
2
31 tháng 7 2023
a: =>(x-2)^3*[(x-2)^8-1]=0
=>(x-2)(x-3)(x-1)=0
=>\(x\in\left\{2;3;1\right\}\)
b: (x-5)^24=(x-5)^9
=>\(\left(x-5\right)^9\cdot\left[\left(x-5\right)^{15}-1\right]=0\)
=>x-5=0 hoặc x-5=1
=>x=6 hoặc x=5
c: =>(x-5)^4*[(x-5)^21-1]=0
=>x-5=0 hoặc x-5=1
=>x=5 hoặc x=6
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
a) \(\left(x-2\right)^{11}=\left(x-2\right)^3\)
\(\Rightarrow\left(x-2\right)^{11}-\left(x-2\right)^3=0\)
\(\Rightarrow\left(x-2\right)^3\left[\left(x-2\right)^8-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^3=0\\\left(x-2\right)^8-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^8=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\left(x-5\right)^{24}=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^{24}-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^9\left[\left(x-5\right)^{15}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^9=0\\\left(x-5\right)^{15}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{15}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
c) \(\left(x-5\right)^{25}=\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^{25}-\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^{21}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\\left(x-5\right)^{21}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{21}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
19 tháng 7 2021
a, Ta có :
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)
\(\Rightarrow x=11;y=17;z=23\)
b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)
\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)
\(\Rightarrow x=6;y=9;z=15\)
TG
19 tháng 7 2021
a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)
Áp dụng t/c dtsbn:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)
b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
xyz = 810
=> 2k.3k.5k = 810
=> k = 3
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)
A )x*2+x*3=50
X*(2+3)=50
X*5=50
X=50:5
X=10
B )x*9-x*5=24
X*(9-5)=24
X*4=24
X=24:4
X=6
A)
x * 2 + x * 3 = 50
x * ( 2 + 3 ) = 50
x * 5 = 50
x = 50 : 5
x = 10
B)
x * 9 - x * 5 = 24
x * ( 9 - 5 ) = 24
x * 4 = 24
x = 24 : 4
x = 6