b4: tìm x
a) x+25+|-x+3|
b)|x+4|+|x-y+10|<0
c)|x-8|+|x+2|=2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x + 10 = 20
<=> x = 20 - 10 = 10
Vậy x = 10
b) 2x + 15 = 35
<=> 2x = 35 - 15 = 20
<=> x = 10
Vậy x = 10
c) 3(x + 2) = 15
<=> x + 2 = 15 : 3 = 5
<=> x = 5 - 2 = 3
Vậy x = 3
d) 10x + 15.11 = 20.10
<=> 10x + 165 = 200
<=> 10x = 200 - 165 = 35
<=> x = 35 : 10 = 3,5
Vậy x = 3,5
e) 4(x + 2) = 3.4
<=> x + 2 = 3
<=> x = 3 - 2 = 1
Vậy x = 1
f) 33x + 135 = 26.9
<=> 33x + 135 = 234
<=> 33x = 234 - 135 = 99
<=> x = 99 : 33 = 3
Vậy x = 3
g) 2x + 15 + 16 + 17 = 100
<=> 2x + 48 = 100
<=> 2x = 100 - 48 = 52
<=> x = 52 : 2 = 26
Vậy x = 26
h) 2(x + 9 + 10 + 11) = 4.12.5
<=> x + 30 = 120
<=> x = 120 - 30 = 90
Vậy x = 90
B2: tính:
a, 225-228:[4.(48-72)]
=225-228:[4.(-24)]
=225-228:(-96)
=225-(-19/8)
=1819/8
5 + ( x + 27 ) = 64
( x + 27 ) = 64 - 5 ( x + 27 ) = 59 x = 59 - 27 x = 32a) \(\Rightarrow x+27=59\Rightarrow x=32\)
b) \(\Rightarrow x-2=39\Rightarrow x=41\)
c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)
d) \(\Rightarrow5x=35\Rightarrow x=7\)
e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)
f) \(\Rightarrow15+x=37\Rightarrow x=22\)
g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)
h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)
a)
Ta có : vì|1/2-1/3+x| lớn hơn hoặc bằng 0
Còn -1/4-|y| bé hơn hoặc bằng 0
=> ko tồn tại x
b)
Ta có: |x-y| lớn hơn hoặc bằng 0 và|y+9/25| lớn hơn hoặc bằng 0 mà:
| x-y|+ |y+9/25| =0 => |x-y| =0 và |y+9/25|=0
Xét |y+9/25| có:
| y+9/25|=0 => y+9/25=0 => y=-9/25
Thay y = -9/25 vào |x-y| =0 => x=-9/25
Vậy x=y=-9/25
bài 1 tính giá trị biểu thức
( - 25 ) nhân ( -3 ) nhân x với x = 4
\(\left(-25\right).\left(-3\right).4\)
\(=\left(-25\right).4.\left(-3\right)\)
\(=-100.\left(-3\right)=300\)
( -1 ) nhân ( -4 ) nhân 5 nhân 8 nhân y với y =25
\(\left(-1\right).\left(-4\right).5.8.25\)
\(=4.5.8.25=4.25.5.8\)
\(=100.40=40000\)
( 2ab mũ 2 ) : c với a =4 ; b= -6 ; c =12
\(\left(2.4.\left(-6\right)\right)^2:12\)
\(=\left(-48\right)^2:12\)
\(=2304:12=192\)
[ ( -25 ) nhân ( - 27 ) nhân ( -x ) ] : y với x = 4 ; y = -9
\(\left[\left(-25\right).\left(-27\right).\left(-4\right)\right]:-9\)
\(=-2700:\left(-9\right)\)
\(=300\)
(a mũ 2 _ b mũ 2) : ( a + b ) nhân ( a _ b ) với a + 5 , b = -3
\(\left(5^2-\left(-3\right)^2\right):\left(5-3\right).\left(5+3\right)\)
\(=16:2.8\)
\(=8.8=64\)
\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)
\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)
2: Tìm x
a) Ta có: x+25=40
nên x=40-25=15
Vậy: x=15
b) Ta có: 198-(x+4)=120
\(\Leftrightarrow x+4=198-120=78\)
hay x=78-4=74
Vậy: x=74
c) Ta có: \(\left(2x-7\right)\cdot3=125\)
\(\Leftrightarrow2x-7=\dfrac{125}{3}\)
\(\Leftrightarrow2x=\dfrac{125}{3}+7=\dfrac{125}{3}+\dfrac{21}{3}=\dfrac{146}{3}\)
\(\Leftrightarrow x=\dfrac{146}{3}:2=\dfrac{146}{6}=\dfrac{73}{3}\)
Vậy: \(x=\dfrac{73}{3}\)
d) Ta có: \(x+16⋮x+1\)
\(\Leftrightarrow x+1+15⋮x+1\)
mà \(x+1⋮x+1\)
nên \(15⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(15\right)\)
\(\Leftrightarrow x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)
hay \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)
Vậy: \(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)
\(a,x+25=40\\ \Rightarrow x=40-25\\ \Rightarrow x=15\\ b,198-\left(x+4\right)=120\\ \Rightarrow-\left(x+4\right)=120-198\\ \Rightarrow-\left(x+4\right)=-78\\ \Rightarrow x+4=78\\ \Rightarrow x=78-4\\ \Rightarrow x=74\\ c,\left(2x-7\right).3=125\\ \Rightarrow2x-7=\dfrac{125}{3}\\ \Rightarrow2x=\dfrac{125}{3}+7\\ \Rightarrow2x=\dfrac{146}{3}\\ \Rightarrow x=\dfrac{146}{3}:2\Rightarrow x=\dfrac{73}{3}\\ d,\left(x+16\right)⋮\left(x+1\right)\\ \Rightarrow\left[\left(x+1\right)+15\right]⋮\left(x+1\right)\\ mà:\left(x+1\right)⋮\left(x+1\right)\\ \Rightarrow15⋮\left(x+1\right)\\ \Rightarrow\left(x+1\right)\inƯ\left(15\right)\\ \Rightarrow\left(x+1\right)\in\left\{-15;-1;1;15\right\}\\ \Rightarrow x\in\left\{-16;-2;0;14\right\}\)
Tự kết luận nhé bạn