a) | 5 - x | - 1 = 4
b )
x + 3x = 44 : 4 - 42
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a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
a,\(\frac{1}{5}\times\frac{2}{9}\div\frac{1}{15}\)
\(=\frac{1}{5}\times\frac{2}{9}\times15\)
\(=\left(\frac{1}{5}\times15\right)\times\frac{2}{9}\)
\(=3\times\frac{2}{9}\)
\(=\frac{2}{3}\)
b\(\frac{4}{15}\times\frac{7}{15}\times\frac{5}{4}\)
\(=\left(\frac{4}{15}\times\frac{5}{4}\right)\times\frac{7}{15}\)
\(=\frac{1}{3}\times\frac{7}{15}\)
\(=\frac{7}{45}\)
c,\(\frac{21}{23}\times\frac{5}{11}\times\frac{44}{?}\)
d,\(26\times\frac{13}{42}\div13\)
\(=\left(26\div13\right)\times\frac{13}{42}\)
\(=2\times\frac{13}{42}\)
\(=\frac{13}{21}\)
a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
c: \(=\dfrac{x+2}{x\left(x-2\right)}+\dfrac{2}{x\left(x+2\right)}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x^2+2x+2x-4+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+7x-2}{x\left(x-2\right)\left(x+2\right)}\)
a,
\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b,
\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)
a) 7(x - 5) + 2 = 51
\(\Leftrightarrow\) 7(x - 5) = 51 - 2
\(\Leftrightarrow\) 7(x - 5) = 49
\(\Leftrightarrow\) x - 5 = 49 : 7
\(\Leftrightarrow\) x - 5 = 7
\(\Leftrightarrow\) x = 7 + 5
\(\Leftrightarrow\) x = 12.
Vậy x = 12.
k) 2412 : (3x + 147) = |-38| + (-26)
\(\Leftrightarrow\) 2412 : (3x + 147)= 38 + (-26)
\(\Leftrightarrow\) 2412 : (3x + 147)= 12
\(\Leftrightarrow\) 3x + 147 = 2412 : 12
\(\Leftrightarrow\) 3x + 147 = 201
\(\Leftrightarrow\) 3x = 201 - 147
\(\Leftrightarrow\) 3x = 54
\(\Leftrightarrow\) x = 54 : 3
\(\Leftrightarrow\) x = 18.
Vậy x = 18.
I) 4824 : (4x + 137) = |-59| + (-35)
\(\Leftrightarrow\) 4824 :(4x + 137) = 59 + (-35)
\(\Leftrightarrow\) 4824 :(4x + 137) = 24
\(\Leftrightarrow\) 4x + 137 = 4824 : 24
\(\Leftrightarrow\) 4x + 137 = 201
\(\Leftrightarrow\) 4x = 201 - 137
\(\Leftrightarrow\) 4x = 64
\(\Leftrightarrow\) x = 64 : 4
\(\Leftrightarrow\) x = 16.
Vậy x = 16.
c) |-123| - 5(x - 3) = (-28) + 66
\(\Leftrightarrow\) 123 - 5(x - 3) = 38
\(\Leftrightarrow\) 5(x - 3) = 123 - 38
\(\Leftrightarrow\) 5(x - 3) = 85
\(\Leftrightarrow\) x - 3 = 85 : 5
\(\Leftrightarrow\) x - 3 = 17
\(\Leftrightarrow\) x = 17 + 3
\(\Leftrightarrow\) x = 20.
Vậy x = 20.
m) 7x - 4 . 6 = 2058
\(\Leftrightarrow\) 7x - 24 = 2058
\(\Leftrightarrow\) 7x = 2058 + 24
\(\Leftrightarrow\) 7x = 2082
\(\Leftrightarrow\) x = 2082 : 7
\(\Leftrightarrow\) x = \(\frac{2058}{7}\).
Vậy x = \(\frac{2058}{7}\).
Phần b) ra phân số nên mình để mai làm và cả những bài còn lại nữa.
Chúc bạn học tốt !!!
---------------NHANH NHA MK ĐANG CẦN GẤP--------------
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a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)
=>7/6x=2/3
hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)
hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
a: Khi x>4 thì A+3x-5-x+4=2x-1
b: A=2016
=>3x-5-|x-4|=2016(1)
Trường hợp x>=4
=>2x-1=2016
hay x=2017/2(nhận)
Trường hợp 2: x<4
=>3x-5-(4-x)=2016
=>3x-5-4+x=2016
=>4x-9=2016
hay x=2025/4(loại)
a. Khi x > 4
\(A=3x-5-x-4\)
\(=2x-9\)
b. Ta có A = 2016
\(\Rightarrow3x-5-\left|x-4\right|=2016\)
\(\Leftrightarrow-\left|x-4\right|=2016-3x+5\)
\(\Leftrightarrow\left|x-4\right|=3x-2021\)
TH1: \(\left|x-4\right|\ge0\) khi \(x\ge4\)
\(x-4=3x-2021\)
\(\Leftrightarrow-2x=-2017\Leftrightarrow x=\dfrac{2017}{2}\left(tmđk\right)\)
TH2 : \(\left|x-4\right|< 0\) khi \(x< 4\)
\(x-4=2021-3x\)
\(\Leftrightarrow4x=2025\Leftrightarrow x=\dfrac{2025}{4}\left(ktmđk\right)\)
Vậy : Phương trình có tập nghiệm \(S=\left\{\dfrac{2017}{2}\right\}\)
c:
Trường hợp 1: x<-3
\(\Leftrightarrow-x-3-x-1=3x\)
\(\Leftrightarrow-5x=4\)
hay \(x=-\dfrac{4}{5}\left(loại\right)\)
Trường hợp 2: -3<=x<-1
\(\Leftrightarrow x+3-x-1=3x\)
hay \(x=\dfrac{2}{3}\left(loại\right)\)
Trường hợp 3: x>=-1
\(\Leftrightarrow2x+4=3x\)
hay x=4(nhận)
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)