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Kết quả dương thì trị tuyệt đối dữ nguyên, âm thì đổi dấu @@, cứ thế mà làm
Vd câu a: x>=-1 => x+1 >=0 => |x+1| +x = x+1 +x = 2x+1.
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1+1/A+1/a2+1/a3+1+.../an+1
=1(1/A/a2/a3/...an)
=1.(1/a1+2+3+...+n)
=1.(1/a6+...+n)
=a6+...+n
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\(B=\frac{2929-101}{2.1919+404}\)
\(B=\frac{29.101-101}{38.101+4.101}\)
\(B=\frac{101.\left(29-1\right)}{101.\left(38+4\right)}\)
\(B=\frac{2}{3}\)
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\(\frac{10.\left(4^6.9^5+6^9.120\right)}{8^4.3^{12}-6^{11}}\)
=\(\frac{2.5.\left[\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5\right]}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
=\(\frac{2^{13}.5.3^{10}+2^{13}.5^2.3^{10}}{2^{12}.3^{12}-3^{11}.2^{11}}\)
=\(\frac{2^{13}.5.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
=\(\frac{4.5.6}{3.5}\)
= 8
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a)\(x\ge-7\Leftrightarrow\left|7+x\right|=7+x\)
\(\Leftrightarrow B=\left|7+x\right|-\left(x-8\right)=7+x-x+8=15\)
b)\(C=\left|x+3\right|+\left|x-4\right|\)
TH1: \(x\le-3\)
<=>\(C=\left|x+3\right|+\left|x-4\right|=-x-3+4-x=1-2x\)
TH2: \(-3< x\le4\)
<=>\(C=\left|x+3\right|+\left|x-4\right|=x+3+4-x=7\)
TH3: x > 4
<=>\(C=\left|x+3\right|+\left|x-4\right|=x+3+x-4=2x-1\)
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+) \(\frac{-180}{270}=\frac{-2}{3}\) +) \(\frac{22}{-165}=\frac{-2}{15}\)
+) \(\frac{-39}{261}=\frac{-13}{87}\) +) \(\frac{4.26}{52.20}=\frac{4.13.2}{4.13.2.10}=\frac{1}{10}\)
+) \(\frac{7.\left(-32\right)}{16.35}=\frac{7.\left(-2\right).16}{16.7.5}=\frac{-2}{5}\)
Học tốt
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x + 22 + (-14) = x + [22 + (-14)]
= x + (22 -14) = x + 8