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\(a.\)\(\frac{13x-16}{15}+\frac{x-32}{35}< \frac{x-6}{21}\)\(MC:105\)
\(\Leftrightarrow\frac{7\left(13x-16\right)}{105}+\frac{3\left(x-2\right)}{105}< \frac{5\left(x-6\right)}{105}\)
\(\text{Khử mẫu ta dc pt tương đương vs pt:}\)
\(\Leftrightarrow7\left(13x-16\right)+3\left(x-2\right)< 5\left(x-6\right)\)
\(\Leftrightarrow91x-112+3x-6< 5x-30\)
\(\Leftrightarrow94x-118< 5x-30\)
\(\Leftrightarrow94x-5x< 118-30\)
\(\Leftrightarrow89x< 88\)
\(\Leftrightarrow x< \frac{88}{89}\)
.\(b.\)\(\frac{5x+12}{14}+\frac{11x+28}{3}>\frac{4x+9}{17}\)\(MC:714\)
\(\text{Khi khử mẫu pt ta dc pt tương đương}:\):
\(\Leftrightarrow51\left(5x+12\right)+238\left(11x+28\right)>42\left(4x+9\right)\)
\(\Leftrightarrow255x+612+2618x+6664>168x+378\)
\(\Leftrightarrow2873x+7276>168x+378\)
\(\Leftrightarrow2873x-168x>-7276+378\)
\(\Leftrightarrow2705x>-6898\)
\(\Leftrightarrow x>-\frac{6898}{2705}\)
\(x^2+2x+5\)
\(=x^2+2.x.1+1+4\)
\(=\left(x+1\right)^2+4\ge4\)
Min \(=4\Leftrightarrow x+1=0\Rightarrow x=-1\)
\(4x^2-4x-35\) \(=\left(2x\right)^2-2.2x.1+1-36\)
\(=\left(2x-1\right)^2-6^2\)
\(=\left(2x-7\right)\left(2x+5\right)\)
\(18x^2-5x-2\) \(=\left(x-\frac{1}{2}\right)\left(x+\frac{2}{9}\right)\)
\(8x^3-26x^2+13x+5=\) \(8x^3-8x^2-18x^2+18x-5x+5\)
\(=8x^2\left(x-1\right)-18x\left(x-1\right)-5\left(x-1\right)\)
\(=\) \(\left(8x^2-18x-5\right)\left(x-1\right)\)
\(=\left(x-\frac{5}{2}\right)\left(x+\frac{1}{4}\right)\)\(\left(x-1\right)\)
lần sau bạn ghi rõ đề bài nhé
a, \(A=x^2+10x+39=x^2+2.5.x+25+14=\left(x+5\right)^2+14\ge14\)
Dấu ''='' xảy ra khi \(x=-5\)
Vậy GTNN A là 14 khi x = -5
b, \(B=25x^2-70x+1000=\left(5x\right)^2-2.5x.7+49+951\)
\(=\left(5x-7\right)^2+951\ge951\)
Dấu ''='' xảy ra khi \(x=\frac{7}{5}\)
Vậy GTNN B là 951 khi x = 7/5
c, \(C=49x^2+64x+100=\left(7x\right)^2+2.7x.\frac{32}{7}+\frac{1024}{49}+\frac{3876}{49}\)
\(=\left(7x+\frac{32}{7}\right)^2+\frac{3876}{49}\ge\frac{3876}{49}\)
Dấu ''='' xảy ra khi \(x=-\frac{32}{49}\)
Vậy GTNN C là 3876/49 khi x = -32/49
d, \(D=5x^2+13x+41=5\left(x^2+\frac{13}{5}x+\frac{41}{5}\right)\)
\(=5\left(x^2+2.\frac{6,5}{5}.x+\frac{169}{100}+\frac{651}{100}\right)=5\left(x+\frac{6,5}{5}\right)^2+\frac{651}{20}\ge\frac{651}{20}\)
Dấu ''='' xảy ra khi \(x=-\frac{6,5}{2}\)
Vậy GTNN D là 651/20 khi x = -6,5/2
\(x^3-7x^2-13x+91=0\)
\(\Rightarrow x^2\left(x-7\right)-13\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(x^2-13\right)=0\)
\(\Rightarrow\left(x-7\right)\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)
Tìm được \(x\in\left\{7;\sqrt{13};-\sqrt{13}\right\}\)
a ) \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(< =>36x^2-12x-36x^2+27x=30\)
\(< =>-12x+27x=30\)
\(< =>15x=30\)
\(< =>x=2\)
b )
\(x\left(5-2x\right)+2x\left(x-1\right)=15\)
\(< =>5x-2x^2+2x^2-2x=15\)
\(< =>5x-2x=15\)
\(< =>3x=15\)
\(< =>x=5\)
OK K MÌNH NHA
Mik nghĩ nên nhân tất ra r trừ 1 thể:VD: a) 36x^2-12x - 36x^2+27x = 30 -12x+27x = 30 15 x = 30 <=> x = 2 b) Tg tự nha bn Ừm...Mik ms hk l8 nên ko chắc,nếu sai thì đừng trak mik a Chúc bn hk tốt