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Giải:
1) \(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(=\dfrac{-1}{12}-\dfrac{55}{24}\)
\(=\dfrac{-19}{8}\)
2) \(-1,75-\left(\dfrac{-1}{9}-2\dfrac{1}{18}\right)\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+2\dfrac{1}{18}\)
\(=-\dfrac{7}{4}+\dfrac{1}{9}+\dfrac{37}{18}\)
\(=\dfrac{5}{12}\)
3) \(-\dfrac{5}{6}-\left(-\dfrac{3}{8}+\dfrac{1}{10}\right)\)
\(=-\dfrac{5}{6}+\dfrac{3}{8}-\dfrac{1}{10}\)
\(=-\dfrac{67}{120}\)
4) \(\dfrac{2}{5}+\left(-\dfrac{4}{3}\right)+\left(-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{5}-\dfrac{4}{3}-\dfrac{1}{2}\)
\(=-\dfrac{43}{30}\)
5) \(\dfrac{3}{12}-\left(\dfrac{6}{15}-\dfrac{3}{10}\right)\)
\(=\dfrac{3}{12}-\dfrac{6}{15}+\dfrac{3}{10}\)
\(=\dfrac{3}{20}\)
6) \(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)
\(=8\dfrac{5}{11}+3\dfrac{5}{8}-3\dfrac{5}{11}\)
\(=8+\dfrac{5}{11}+3+\dfrac{5}{8}-3-\dfrac{5}{11}\)
\(=8+\dfrac{5}{8}\)
\(=\dfrac{69}{8}\)
7) \(-\dfrac{1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}.13\dfrac{9}{11}-\dfrac{1}{4}.6\dfrac{2}{11}\)
\(=-\dfrac{1}{4}\left(13\dfrac{9}{11}+6\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+\dfrac{9}{11}+6+\dfrac{2}{11}\right)\)
\(=-\dfrac{1}{4}\left(13+6+1\right)\)
\(=-\dfrac{1}{4}.20=-5\)
8) \(\dfrac{4}{9}:\left(-\dfrac{1}{7}\right)+6\dfrac{5}{9}:\left(-\dfrac{1}{7}\right)\)
\(=\dfrac{4}{9}\left(-7\right)+6\dfrac{5}{9}\left(-7\right)\)
\(=-7\left(\dfrac{4}{9}+6\dfrac{5}{9}\right)\)
\(=-7\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)\)
\(=-7\left(6+1\right)\)
\(=-7.7=-49\)
Vậy ...
Bài 1:
a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)
\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)
\(\Leftrightarrow-5=0\)(vl)
Vậy: \(x\in\varnothing\)
b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)
hay x=1
Vậy: x=1
c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)
\(\Leftrightarrow2x-72=0\)
\(\Leftrightarrow2\left(x-36\right)=0\)
mà 2>0
nên x-36=0
hay x=36
Vậy: x=36
d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)
\(\Leftrightarrow120x+36=56-64x\)
\(\Leftrightarrow120x+36-56+64x=0\)
\(\Leftrightarrow184x-20=0\)
\(\Leftrightarrow184x=20\)
hay \(x=\frac{5}{46}\)
Vậy: \(x=\frac{5}{46}\)
e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)
\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)
\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)
\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)
\(\Leftrightarrow-23x+29=0\)
\(\Leftrightarrow-23x=-29\)
hay \(x=\frac{29}{23}\)
Vậy: \(x=\frac{29}{23}\)
f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)
\(\Leftrightarrow2x+8-10x-50-25=0\)
\(\Leftrightarrow-8x-67=0\)
\(\Leftrightarrow-8x=67\)
hay \(x=\frac{-67}{8}\)
Vậy: \(x=\frac{-67}{8}\)
g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)
\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)
\(\Leftrightarrow10-5x-8x-8+12x-30=0\)
\(\Leftrightarrow-x-28=0\)
\(\Leftrightarrow-x=28\)
hay x=-28
Vậy: x=-28
h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)
\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(x\in R\)
Bài 2:
a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)
b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)
c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)
\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: Tập nghiệm S={-3}
d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)
\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)
\(\Leftrightarrow12-7x=0\)
\(\Leftrightarrow7x=12\)
hay \(x=\frac{12}{7}\)
Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)
e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x
\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow31x=1\)
hay \(x=\frac{1}{31}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)
Bài 11:
1) Sửa lại đề là: \(A=127^2+146.127+73^2\)
\(\Rightarrow A=127^2+2.127.73+73^2\)
\(\Rightarrow A=\left(127+73\right)^2\)
\(\Rightarrow A=200^2\)
\(\Rightarrow A=40000\)
Vậy \(A=40000.\)
2) Sửa lại đề là: \(B=9^8.2^8-\left(18^4-1\right).\left(18^4+1\right)\)
\(\Rightarrow B=\left(9.2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)
\(\Rightarrow B=18^8-\left(18^8-1\right)\)
\(\Rightarrow B=18^8-18^8+1\)
\(\Rightarrow B=0+1\)
\(\Rightarrow B=1\)
Vậy \(B=1.\)
4) \(D=\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(\Rightarrow2D=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
\(\Rightarrow D=\frac{3^{32}-1}{2}\)
a) \(\frac{x+5}{4}\)-\(\frac{2x-5}{3}\)=\(\frac{6x-1}{3}\)+\(\frac{2x-3}{12}\)
⇔\(\frac{3\left(x+5\right)}{12}\)-\(\frac{4\left(2x-5\right)}{12}\)=\(\frac{4\left(6x-1\right)}{12}\)+\(\frac{2x-3}{12}\)
⇒ 3x+15-8x+20=24x-4+2x-3
⇔3x+15-8x+20-24x+4-2x+3=0
⇔-31x+42=0
⇔x=\(\frac{42}{31}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{42}{31}\)}
a: \(=\dfrac{2\cdot5^5-4\cdot5^3+5^4}{5^3}=2\cdot5^2-4+5=50+1=51\)
b: \(=\dfrac{3^8-3^6+3^6\cdot2^3}{3^5}=3^3-3+3\cdot2^3=24+24=48\)
c: \(=\dfrac{7^6\cdot2^3-7^3}{7^3}=14^3-1\)
d: \(=28^4-28^4+1=1\)