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Câu a:
1 + {-2 - [-3 + (-4 + |x|)]} = 1 - 2 + [(-3 - 4)]
1 + {-2 -[-3 - 4 + |x|]} = 1 - 2 + [-7]
1 + {-2 - [- 7 + |x|]} = 1 - 2 - 7
1 + {- 2 + 7 - |x|} = - 1 - 7
1 + {5 - |x|} = - 8
5 - |x| = - 8 - 1
5 - |x| = - 9
|x| = 5 + 9
|x| = 14
x = - 14 hoặc x = 14
Vậy x ∈ {-14; 14}
Câu b:
34 + (9 - 21) = 3417 - (x + 3417)
34 + (-12) = 3417 - x - 3417
34 - 12 = 3417 - 3417 - x
22 = - x
x = 22 : (-1)
x = - 22
Vậy x = - 22
Câu a:
18 - |x - 1| = 2^2
18 - |x - 1| = 4
|x - 1| = 18 - 4
|x - 1| = 14
x - 1 = -14 hoặc x - 1 = 14
Th1: x - 1 = -14
x = - 14 + 1
x = -13
Th2: x - 1 = 14
x = 14 + 1
x = 15
Vậy x ∈ {-13; 15}
Câu b:
x - {55 - [49 + (-28 - x)]} = 13 - {47 + [25 - (32 - x)]}
x - {55 - [49 - 28 - x]} = 13 - {47 + [25 - 32 + x]}
x - {55 - 49 + 28 + x} = 13 - {47 + 25 - 32 + x}
x - {6 + 28 + x} = 13 - {72 - 32 + x}
x - {34 + x} = 13 - {40 + x}
x - 34 - x = 13 - 40 - x
(x - x) + x = 13 - 40 + 34
0 + x = - 27 + 34
x = 7
Vậy x = 7
bài 2) a) \(2\left(x+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\) vậy \(x=-1\)
b) \(x\left(x-2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\) vậy \(x=0;x=2\)
c) \(\left(x-1\right)\left(x+7\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) vậy \(x=1;x=-7\)
d) \(\left(x+2\right)\left(x^2-9\right)=0\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x^2-9=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x^2=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\end{matrix}\right.\) vậy \(x=-2;x=3;x=-3\)
e) \(x^2\left(x-5\right)+2\left(x-5\right)=0\Leftrightarrow\left(x^2+2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2=0\\x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\) vậy \(x=5\)
bài 1) \(A=48+\left(-48-174\right)+\left|-74\right|=48-48-174+74=-100\)
\(B=\left(-123\right)+77+\left(-257\right)-23-43=-123+77-257-23-43=-369\)
\(C=\left(-57\right)+\left(-159\right)+47+169=-57-159+47+169=0\)
quá hợp lí 
g)=>x+1/2=0
x=0-1/2
x=-1/2
hoặc 2/3-2x=0
2x=2/3-0
2x=2/3
x=2/3:2
x=1/3
nhìn @_@ hoa cả mắt đăng từng bài thôi bạn
Bài 1: Tính ( hợp lý nếu có thể )
\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)
\(=-1+1+\dfrac{2}{-5}\)
\(=0+\dfrac{2}{-5}\)
\(=\dfrac{2}{-5}\)
\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)
\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}\)
\(=\dfrac{2}{3}\)
\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1\)
\(=0\)
\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)
\(=\dfrac{1}{6}+\dfrac{-1}{6}\)
\(=0\)
Bài 2: Tìm x,biết:
a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{2}{3}\)
\(x=\dfrac{2}{15}\)
Vậy \(x=\dfrac{2}{15}\)
b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)
\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=\dfrac{3}{3}=1\)
Vậy \(x=1\)
c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!
\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)
\(x=\dfrac{1}{44}\)
Vậy \(x=\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
a: =>15-(x-2)=-13-27=-40
=>x-2=15+40=55
hay x=57
b: =>5-x=-114+12=-102
=>x=107
c: \(\Leftrightarrow\left|x\right|=-1-5=-6\)(vô lý)
d: \(\Leftrightarrow\left|x-3\right|=3\)
=>x-3=3 hoặc x-3=-3
=>x=6 hoặc x=0
|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|
\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)
\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)
\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)
Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}
|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0
|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|
\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)
\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)
\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)
Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}
Câu c)
\(7^{x + 1} + 7^{x} = 8 \times 7^{5}\)
Bước 1: Đặt \(7^{x} = a\)
\(7^{x + 1} = 7 \cdot 7^{x} = 7 a\)
Bước 2: Thay vào phương trình
\(7 a + a = 8 \times 7^{5}\) \(8 a = 8 \times 7^{5}\)
Bước 3: Chia cả hai vế cho 8
\(a = 7^{5}\)
Bước 4: Trả lại \(a = 7^{x}\)
\(7^{x} = 7^{5} \Rightarrow x = 5\)
✅ Kết quả: \(x = 5\)
Câu d)
\(11^{x + 3} + 11^{x + 2} = 12 \times 11^{10}\)
Bước 1: Đặt \(11^{x} = b\)
\(11^{x + 3} = 11^{3} \cdot 11^{x} = 1331 b\) \(11^{x + 2} = 11^{2} \cdot 11^{x} = 121 b\)
Bước 2: Thay vào phương trình
\(1331 b + 121 b = 12 \times 11^{10}\) \(1452 b = 12 \times 11^{10}\)
Bước 3: Chia cả hai vế cho 1452
Trước hết:
\(1452 = 12 \times 121\)
nên:
\(b = \frac{12 \times 11^{10}}{12 \times 121} = \frac{11^{10}}{11^{2}} = 11^{8}\)
Bước 4: Trả lại \(b = 11^{x}\)
\(11^{x} = 11^{8} \Rightarrow x = 8\)
✅ Kết quả: \(x = 8\)
c: \(7^{x+1}+7^{x}=8\cdot7^5\)
=>\(7^{x}\cdot7+7^{x}=8\cdot7^5\)
=>\(8\cdot7^{x}=8\cdot7^5\)
=>\(7^{x}=7^5\)
=>x=5
d: \(11^{x+2}+11^{x+3}=12\cdot11^{10}\)
=>\(11^{x+2}+11^{x+2}\cdot11=12\cdot11^{10}\)
=>\(11^{x+2}\cdot12=11^{10}\cdot12\)
=>\(11^{x+2}=11^{10}\)
=>x+2=10
=>x=8
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.
b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.
c) x : 7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.
d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.
e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.
g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135