diện tích xung quanh HLP:

\(5,4\times5,4\times4=116,64\left(dm^2\right)\)

diện tích toàn phần HLP:

\(5,4\times5,4\times6=174,96\left(dm^2\right)\)

⇒ đáp án A

gọi s(km) là quãng đường AB) (s>0)

vận tốc xe lúc về là: 45 - 10 = 35 (km/h)

thời gian xe đi từ A đến B: \(\frac{s}{45}\left(giờ\right)\)

thời gian xe đi từ B đến A: \(\frac{s}{35}\left(giờ\right)\)

vì thời gian về nhiều hơn thời gian đi 12p nên ta có phương trình:

\(\frac{s}{35}-\frac{s}{45}=\frac{12}{60}\)

\(45s-35s=\frac{12}{60}\cdot35\cdot45\)

\(10s=315\Rightarrow s=31,5\left(TM\right)\)

vậy quãng đường AB dài 31,5km

1.1) a) \(\left|2x-5\right|=4\)

\(\Rightarrow\left[\begin{array}{l}2x-5=4\\ 2x-5=-4\end{array}\Rightarrow\left[\begin{array}{l}2x=9\\ 2x=1\end{array}\Rightarrow\left[\begin{array}{l}x=\frac92\\ x=\frac12\end{array}\right.\right.\right.\)

vậy \(x\in\left\lbrace\frac92;\frac12\right\rbrace\)

b)) \(\frac13-\left|\frac54-2x\right|=\frac14\)

\(\left|\frac54-2x\right|=\frac13-\frac14\)

\(\left|\frac54-2x\right|=\frac{1}{12}\)

\(\Rightarrow\left[\begin{array}{l}\frac54-2x=\frac{1}{12}\\ \frac54-2x=-\frac{1}{12}\end{array}\Rightarrow\left[\begin{array}{l}2x=\frac54-\frac{1}{12}\\ 2x=\frac54-\left(-\frac{1}{12}\right)\end{array}\right.\right.\)

\(\Rightarrow\left[\begin{array}{l}2x=\frac76\\ 2x=\frac43\end{array}\Rightarrow\left[\begin{array}{l}x=\frac{7}{12}\\ x=\frac23\end{array}\right.\right.\)

vậy \(x\in\left\lbrace\frac{7}{12};\frac23\right\rbrace\)

\(c.\frac12-\left|x+\frac15\right|=\frac13\)

\(\left|x+\frac15\right|=\frac12-\frac13\)

\(\left|x+\frac15\right|=\frac16\)

\(\Rightarrow\left[\begin{array}{l}x+\frac15=\frac16\\ x+\frac15=-\frac16\end{array}\Rightarrow\left[\begin{array}{l}x=\frac16-\frac15\\ x=-\frac16-\frac15\end{array}\right.\right.\Rightarrow\left[\begin{array}{l}x=-\frac{1}{30}\\ x=-\frac{11}{30}\end{array}\right.\)

vậy \(x\in\left\lbrace-\frac{1}{30};-\frac{11}{30}\right\rbrace\)

\(d.\frac34-\left|2x+1\right|=\frac78\)

\(\left|2x+1\right|=\frac34-\frac78\)

\(\left|2x+1\right|=-\frac18\)

\(\) ⇒ x thuộc rỗng

1.2) a) \(2\left|2x-3\right|=\frac12\)

\(\left|2x-3\right|=\frac12:2=\frac12\cdot\frac12=\frac14\)

\(\left[\begin{array}{l}2x-3=\frac14\\ 2x-3=-\frac14\end{array}\Rightarrow\left[\begin{array}{l}2x=\frac14+3\\ 2x=-\frac14+3\end{array}\right.\right.\)

\(\left[\begin{array}{l}2x=\frac{13}{4}\\ 2x=\frac{11}{4}\end{array}\Rightarrow\left[\begin{array}{l}x=\frac{13}{4}:2=\frac{13}{4}\cdot\frac12=\frac{13}{8}\\ x=\frac{11}{4}:2=\frac{11}{4}\cdot\frac12=\frac{11}{8}\end{array}\right.\right.\)

vậy: \(x\in\left\lbrace\frac{13}{8};\frac{11}{8}\right\rbrace\)

\(\frac{b)1}{3}-\left|\frac54-2x\right|=\frac14\)

\(\left|\frac54-2x\right|=\frac13-\frac14\)

\(\left|\frac54-2x\right|=\frac{1}{12}\)

\(\left[\begin{array}{l}\frac54-2x=\frac{1}{12}\\ \frac54-2x=-\frac{1}{12}\end{array}\Rightarrow\left[\begin{array}{l}2x=\frac54-\frac{1}{12}\\ 2x=\frac54-\left(-\frac{1}{12}\right)\end{array}\right.\right.\)

\(\left[\begin{array}{l}2x=\frac76\\ 2x=\frac43\end{array}\Rightarrow\left[\begin{array}{l}x=\frac76:2=\frac76\cdot\frac12=\frac{7}{12}\\ x=\frac43:2=\frac43\cdot\frac12=\frac23\end{array}\right.\right.\)

vậy \(x\in\left\lbrace\frac{7}{12};\frac23\right\rbrace\)

\(c.\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)

\(\left|x+\frac{4}{15}\right|=3,75-2,15\)

\(\left|x+\frac{4}{15}\right|=1,6\)

\(\left[\begin{array}{l}x+\frac{4}{15}=1,6\\ x+\frac{4}{15}=-1,6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1,6-\frac{4}{15}\\ x=-1,6-\frac{4}{15}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=-\frac{28}{15}\end{array}\right.\)

vậy \(x\in\left\lbrace\frac43;-\frac{28}{15}\right\rbrace\)

\(\frac{x+1}{x-1}+\frac{x-1}{x+1}=4\) (đkxđ: \(x\ne\pm1\) )

\(\left(x+1\right)^2+\left(x-1\right)^2=4x^2-4\)

\(x^2+2x+1+x^2-2x+1=4x^2-4\)

\(2x^2+2=4x^2-4\)

\(4x^2-2x^2=4+2\)

\(2x^2=6\)

\(x^2=3\Rightarrow x=\pm\sqrt{3}\left(TM\right)\)

\(1)\left(2x+1\right)^2+\left(2x-1\right)^2\)

\(=4x^2+4x+1+4x^2-4x+1\)

\(=8x^2+2\)

\(2)-\left(x+1\right)^2-\left(x-1\right)^2\)

\(=-\left\lbrack\left(x+1\right)^2+\left(x-1\right)^2\right\rbrack\)

\(=-\left\lbrack\left(x^2+2x+1\right)+\left(x^2-2x+1\right)\right\rbrack\)

\(=-\left\lbrack2x^2+2\right\rbrack=-2x^2-2\)

\(3)\left(x+2y\right)^2-\left(x-2y\right)^2\)

\(=\left(x+2y+x-2y\right)\left(x+2y-x+2y\right)\)

\(=2x\cdot4y=8xy\)

\(4)\left(3x+y\right)^2+\left(x-y\right)^2\)

\(=9x^2+6xy+y^2+x^2-2xy+y^2\)

\(=10x^2+4xy+2y^2\)

\(5)-\left(x+5\right)^2-\left(x-3\right)^2\)

\(=-\left\lbrack\left(x+5\right)^2+\left(x-3\right)^2\right\rbrack\)

\(=-\left\lbrack x^2+10x+25+x^2-6x+9\right\rbrack\)

\(=-\left\lbrack2x^2+4x+34\right\rbrack=-2x^2-4x-34\)

\(6)\left(3x-2\right)^2-\left(3x-1\right)^2\)

\(=\left(3x-2+3x-1\right)\left(3x-2-3x+1\right)\)

\(=\left(6x-3\right)\cdot\left(-1\right)=-6x+3\)

\(7)\left(x-4y\right)^2+\left(x+4y\right)^2\)

\(=x^2-8xy+16y^2+x^2+8xy+16y^2\)

\(=2x^2+32y^2\)

\(8)-\left(-2x+3\right)^2-\left(5x-3\right)^2\)

\(=-\left\lbrack\left(4x^2-12x+9)+\left(25x^2-30x+9\right)\right\rbrack\right.\)

\(=-29x^2+42x-18\)

\(9)\left(-2x+3\right)^2-\left(5x-3\right)^2\)

\(=\left(-2x+3+5x-3\right)\left(-2x+3-5x+3\right)\)

\(=3x\cdot\left(-7x+6\right)\)

\(=-21x^2+18x\)

\(a.x:\left(-\frac23\right)-\frac12\left|+\frac56\right|\cdot\frac12=\frac34\)

\(x\cdot\left(-\frac32\right)-\frac12+\frac{5}{12}=\frac34\)

\(x\cdot\left(-\frac32\right)=\frac34-\frac{5}{12}+\frac12\)

\(x\cdot\left(-\frac32\right)=\frac56\)

\(x=\frac56:\left(-\frac32\right)=\frac56\cdot\left(-\frac23\right)\)

\(x=-\frac59\)

\(b.\left(-\frac23\right)x+\frac38\cdot\left(-\frac85\right)=-\frac{8}{15}\)

\(\left(-\frac23\right)x-\frac35=-\frac{8}{15}\)

\(\left(-\frac23\right)x=-\frac{8}{15}+\frac35=\frac{1}{15}\)

\(x=\frac{1}{15}:\left(-\frac23\right)=\frac{1}{15}\cdot\left(-\frac32\right)\)

\(x=-\frac{1}{10}\)

\(x^2-4=\left(x-2\right)\left(x+2\right)\)

\(\frac{2}{1\times2+}\frac{2}{2\times3}+\frac{2}{3\times4}+\cdots+\frac{2}{98\times99}+\frac{2}{99\times100}\)

\(=2\cdot\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\cdots+\frac{1}{99\times100}\right)\)

\(=2\cdot\left(\frac11-\frac12+\frac12-\frac13+\cdots+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\cdot\left(\frac11-\frac{1}{100}\right)=2\cdot\frac{99}{100}=\frac{99}{50}\)

\(G=\frac{4x-1}{2x+3}=\frac{2\cdot\left(2x+3\right)-6-1}{2x+3}\)

\(=\frac{2\cdot\left(2x+3\right)-7}{2x+3}=2-\frac{7}{2x+3}\)

để G là số nguyên thì 2x+3 thuộc Ư(7) = \(\left\lbrace\pm1;\pm7\right\rbrace\)

vì x thuộc N nên \(x\ge0\)

\(\Rightarrow2x\ge0\Rightarrow2x+3\ge3\)

ta có: 2x + 3 = 7

⇒ 2x = 7 - 3 = 4

⇒ x = 4 : 2 = 2

vậy để G nhận giá trị nguyên thì x = 2

\(a.97-7\left(x-8\right)=3\cdot2^4\)

\(97-7\left(x-8\right)=48\)

\(7\left(x-8\right)=97-48\)

\(7\left(x-8\right)=49\)

\(x-8=49:7=7\)

\(x=7+8=15\)

\(b.5x^2-74=51\)

\(5x^2=51+74=125\)

\(x^2=125:5=25\)

\(\Rightarrow x=\pm5\)

\(c.3x+2-3x=216\)

\(0x=214\)

⇒ \(x\in\) rỗng

\(5\left(3x-4\right)^3=40\)

\(\left(3x-4\right)^3=40:5=8\)

⇒ 3x - 4 = 2

⇒ 3x = 2 + 4 = 6

⇒ x = 6 : 3 = 2