Tìm x biết:
a) 2 3/11 - x = 5/44
b) x : 2/7 = 34,5 - 6,5
1) Thực hiện phép tính:
a) (-3+4):2 b) -4+(-5)-17 c) 4-[(-11)-(2+5)]
5 9 7 9 9 4 3 6 9 3
2) Tìm x biết:
a) 2+7:x=5 b) (-0,75 x +5).4-(-1)=-5 c)-x+5=1,2 d)2+2x=1
3 4 6 2 7 3 6 3 3 2
3) Tìm số nguyên x,y biết:
x-1=1
4 y 2
Bạn ơi, bạn viết lại đề đi. Khó nhìn quá
Tìm x, biết:
a)\(x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2};\) b)\(x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9};\)
c)\({\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9};\) d)\(x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\)
a)
\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} = - \frac{1}{2}\\x = - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
b)
\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)
Vậy \(x = \frac{9}{{25}}\).
c)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)
Vậy \(x = \frac{4}{9}\).
d)
\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)
Vậy \(x = \frac{1}{{16}}\).
Tìm x, biết:
a)\(x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\);
b)\(3,7 - x = \frac{7}{{10}};\)
c)\(x.\frac{3}{2} = 2,4\);
d)\(3,2:x = - \frac{6}{{11}}\).
a)
\(\begin{array}{l}x + \left( { - \frac{1}{5}} \right) = \frac{{ - 4}}{{15}}\\x = \frac{{ - 4}}{{15}} + \frac{1}{5}\\x = \frac{{ - 4}}{{15}} + \frac{3}{{15}}\\x = \frac{{ - 1}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 1}}{{15}}\).
b)
\(\begin{array}{l}3,7 - x = \frac{7}{{10}}\\x = 3,7 - \frac{7}{{10}}\\x = \frac{{37}}{{10}} - \frac{7}{{10}}\\x=\frac{30}{10}\\x = 3\end{array}\)
Vậy \(x = 3\).
c)
\(\begin{array}{l}x.\frac{3}{2} = 2,4\\x.\frac{3}{2} = \frac{{12}}{5}\\x = \frac{{12}}{5}:\frac{3}{2}\\x = \frac{{12}}{5}.\frac{2}{3}\\x = \frac{8}{5}\end{array}\)
Vậy \(x = \frac{8}{5}\)
d)
\(\begin{array}{l}3,2:x = - \frac{6}{{11}}\\\frac{{16}}{5}:x = - \frac{6}{{11}}\\x = \frac{{16}}{5}:\left( { - \frac{6}{{11}}} \right)\\x = \frac{{16}}{5}.\frac{{ - 11}}{6}\\x = \frac{{ - 88}}{{15}}\end{array}\)
Vậy \(x = \frac{{ - 88}}{{15}}\).
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
Tìm x biết:
a, \(\dfrac{3}{5}:x+\dfrac{1}{5}=\dfrac{11}{25}\)
b, \(2\left(x-\dfrac{1}{3}\right)-1\dfrac{2}{3}=\dfrac{-23}{15}\)
c, \(\left|x+1\right|-\dfrac{1}{7}=\dfrac{1}{3}\)
d, \(\dfrac{x+1}{3}=\dfrac{2x-1}{5}\)
a/ => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)
=> \(\dfrac{1}{x}=\dfrac{2}{5}\)
=> x = 5/2
b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)
=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)
=> \(x=\dfrac{2}{5}\)
c/ => | x + 1| = 10/21
=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)
d/ => \(5x+5=6x-3\)
=> x = 8
Bài 2: Tìm x, biết:
a) 75 - (x + 11) = 13
b) 29 + (x + 11) = 57
c) 11 + x : 5 = 13
d) 13 + 2(x + 1) = 15
e) 2x + 21 = 41
f) 12 + 3(x – 2) = 60
g) 24x – 11.13 = 11.11
h) 17 – (x – 4) : 2 = 3
c: Ta có: 11+x:5=13
\(\Leftrightarrow x:5=2\)
hay x=10
d: Ta có: \(13+2\left(x+1\right)=15\)
\(\Leftrightarrow2x+2=2\)
\(\Leftrightarrow2x=0\)
hay x=0
e: Ta có: 2x+21=41
\(\Leftrightarrow2x=20\)
hay x=10
f: Ta có: \(12+3\left(x-2\right)=60\)
\(\Leftrightarrow3\left(x-2\right)=48\)
\(\Leftrightarrow x-2=16\)
hay x=18
g: Ta có: \(24x-11\cdot13=11\cdot11\)
\(\Leftrightarrow24x=11\cdot24\)
hay x=11
h: Ta có: \(17-\left(x-4\right):2=3\)
\(\Leftrightarrow\left(x-4\right):2=14\)
\(\Leftrightarrow x-4=28\)
hay x=32
Tìm x, biết:
a) (2x-1)2+(x+3)2-5(x+7)(x-7)=0
b) x(x-5)(x+5)-(x+2)(x2-2x+4)=3
giúp tui với
\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)
\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)
Vậy \(x=\dfrac{-255}{2}\)
\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)
Câu 2: Tìm x biết:
a)2/9 + x = 1 b) x : 7/11= 22
a x = 1 - 2/9
x = 7/9
b x = 22 x 7/11
x = 14
a, \(\Leftrightarrow x=1-\dfrac{2}{9}=\dfrac{7}{9}\)
b, \(\Leftrightarrow x=\dfrac{22.7}{11}=2.7=14\)
Câu 2: Tìm x biết:
a)2/9 + x = 1
=1-2/9
=9/9-2/9
=>7/9
b) x : 7/11= 22
=22*7/11
=14
Tìm x biết:a) 10/x = -15/9; b) x/9 = -11/5 : 0,6; c) -7/8 - 2x = -3/4; d) (x-1/2):1/3+5/7=9và5/7; e)1/15.x+4/5.x=5và1/5; f) -1/3<x/6<1/2(x thuộc Z); h) 3/5+2/5:x=-1/4;i) 4 và 3/4x - 3 và 1/2=5/4; k)9/4.(1/3x-1/2)=4và1/2