Tìm x
a)12 - |x| = 8
b) (x-2/3)2 25/16
c) (3/4-x)3 =25/16
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c: Ta có: \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x=0\)
\(\Leftrightarrow x\left(3x+26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\)
\(a,\Leftrightarrow x^2+8x+16-x^3-12x^2=16\\ \Leftrightarrow x^3+11x^2-8x=0\\ \Leftrightarrow x\left(x^2+11x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+11x-8=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=121+32=153\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11-3\sqrt{17}}{2}\\x=\dfrac{-11+3\sqrt{17}}{2}\end{matrix}\right.\\ S=\left\{0;\dfrac{-11-3\sqrt{17}}{2};\dfrac{-11+3\sqrt{17}}{2}\right\}\)
\(c,\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\\ \Leftrightarrow3x^2+26x=0\\ \Leftrightarrow x\left(3x+26\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{26}{3}\end{matrix}\right.\\ d,\Leftrightarrow x^3-6x^2+12x-8-x^3-125-6x^2=11\\ \Leftrightarrow-12x^2+12x-144=0\\ \Leftrightarrow x^2-x+12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
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
Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
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
Bài 3:
Gọi số nhóm là x
Theo đề, ta có: \(x\in\left\{1;2;3;4;6;9;12;18;36\right\}\)
mà 2<x<6
nên \(x\in\left\{3;4\right\}\)
Vậy: Có 2 cách chia nhóm
a) 25 - x = 12 + 6 =18
x=25-18=7 Vậy x=7
b) 7 + 2 x ( x -3 ) = 11
2.(x-3)=11-7=4
x-3=4:2=2
x=3+2=5
c) 102 : ( 2.x + 13) : 4) = 6
(2.x+13):4=102:6=17
2.x+13=17.4=68
2.x=68-13=55
x=27,5 Vậy x=27,5
Bài 3:
Gọi số nhóm là x
Theo đề, ta có: x∈{1;2;3;4;6;9;12;18;36}x∈{1;2;3;4;6;9;12;18;36}
mà 2<x<6
nên x∈{3;4}x∈{3;4}
Vậy: Có 2 cách chia nhóm
còn bài 1 chắc bn làm đc nha tick mk nha
Bài 3:
Tổng số tiền An dùng mua đồ:
125 000 + 85 000 + 60 000 + 65 000 = 335 000 (đồng)
Số tiền An còn lại sau khi mua đồ:
350 000 - 335 000 = 15 000 (đồng)
Đ.số: 15 000 đồng
Bài 2:
a, IV: Bốn(4); XXVII: Hai mươi bảy (27), XXX: ba mươi (30), M:một nghìn (1000)
b, 7: VII; 15: XV; 29: XXIX
\(a,x+\dfrac{2}{5}=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}-\dfrac{2}{5}\)
\(x=\dfrac{5}{10}-\dfrac{4}{10}\)
\(\Rightarrow x=....\)
\(a,x+\dfrac{2}{5}=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{2}-\dfrac{2}{5}=\dfrac{5-4}{10}=\dfrac{1}{10}\)
\(b,x-\dfrac{2}{5}=\dfrac{1}{7}\\ \Rightarrow x=\dfrac{1}{7}+\dfrac{2}{5}=\dfrac{5+14}{35}=\dfrac{19}{35}\)
\(c,x\cdot\dfrac{3}{4}=\dfrac{9}{20}\\ \Rightarrow x=\dfrac{9}{20}:\dfrac{3}{4}=\dfrac{9}{20}\cdot\dfrac{4}{3}=\dfrac{3\cdot1}{5\cdot1}=\dfrac{3}{5}\)
\(d,x:\dfrac{1}{7}=14\\ \Rightarrow x=14\cdot\dfrac{1}{7}=\dfrac{14}{7}=2\)
\(e,\dfrac{2}{3}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{2}{3}-\dfrac{1}{5}=\dfrac{10-3}{15}=\dfrac{7}{15}\)
\(f,\dfrac{4}{15}:x=\dfrac{12}{25}\\ \Rightarrow x=\dfrac{4}{15}:\dfrac{12}{25}=\dfrac{4}{15}\cdot\dfrac{25}{12}=\dfrac{1\cdot5}{3\cdot3}=\dfrac{5}{9}\)
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
12-|x|=8
=> |x|= 12-8= 4
=> x= 4 hoặc x= -4
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