\(A=3\left|1-2x\right|-5\)

Ta có: \(\left|1-2x\right|\ge0\forall x\)

\(\Rightarrow3.\left|1-2x\right|-5\ge-5\forall x\)

\(\Rightarrow A\ge-5\forall x\)

Dấu "=" xảy ra

\(\Leftrightarrow3.\left|1-2x\right|=0\Leftrightarrow1-2x=0\Leftrightarrow x=\dfrac{1}{2}\)

thank bn

a: 5-3x=6x+7

=>-3x-6x=7-5

=>-9x=2

=>\(x=-\dfrac{2}{9}\)

b: \(\dfrac{3x-2}{6}-5=3-\dfrac{2\left(x+7\right)}{4}\)

=>\(\dfrac{3x-2}{6}+\dfrac{x+7}{2}=8\)

=>\(\dfrac{3x-2+3\left(x+7\right)}{6}=8\)

=>3x-2+3x+14=48

=>6x+12=48

=>6x=36

=>\(x=\dfrac{36}{6}=6\)

c: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

=>\(\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)

=>(x-1)(5x+3-3x+8)=0

=>(x-1)(2x+11)=0

=>\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)

d: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

=>\(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)

=>\(\left(x-4\right)\left(3x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

a: Ta có: ΔABC cân tại A

mà AH là đường cao

nên H là trung điểm của BC

=>\(BH=CH=\dfrac{6}{2}=3\left(cm\right)\)

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HA^2=5^2-3^2=16\)

=>\(HA=\sqrt{16}=4\left(cm\right)\)

b: Xét ΔAHB có HE là phân giác

nên \(\dfrac{AE}{EB}=\dfrac{AH}{HB}=\dfrac{4}{3}\)(1)

=>\(\dfrac{AE}{4}=\dfrac{EB}{3}\)

mà AE+EB=AB=5cm

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{AE}{4}=\dfrac{EB}{3}=\dfrac{AE+EB}{4+3}=\dfrac{5}{7}\)

=>\(AE=\dfrac{5}{7}\cdot4=\dfrac{20}{7}\left(cm\right)\)

c: Xét ΔAHC có HF là phân giác

nên \(\dfrac{AF}{FC}=\dfrac{AH}{HC}=\dfrac{4}{3}\left(2\right)\)

Từ (1) và (2) suy ra \(\dfrac{AE}{EB}=\dfrac{AF}{FC}\)

Xét ΔABC có \(\dfrac{AE}{EB}=\dfrac{AF}{FC}\)

nên EF//BC

Ta có: EF//BC

BC\(\perp\)AH

Do đó: EF\(\perp\)AH

d: Xét ΔAHB vuông tại H có HE là đường cao

nên \(HE\cdot AB=HA\cdot HB\)

=>\(HE\cdot5=3\cdot4=12\)

=>\(HE=\dfrac{12}{5}=2,4\left(cm\right)\)

Xét ΔABC có EF//BC

nên \(\dfrac{EF}{BC}=\dfrac{AE}{AB}\)

=>\(\dfrac{EF}{6}=\dfrac{20}{7}:5=\dfrac{4}{7}\)

=>\(EF=\dfrac{24}{7}\left(cm\right)\)

Chọn D

Có phải bạn đang thi ko?

Đây là bài ôn thi 4h là bắt buộc phải nộp

Để ΔA'B'C'~ΔABC thì \(\widehat{B}=\widehat{B'}\) hoặc \(\widehat{C}=\widehat{C'}\) hoặc \(\widehat{B}=\widehat{C'}\) hoặc \(\widehat{B'}=\widehat{C}\)

Khi đó, hai tam giác đồng dạng với nhau theo trường hợp g-g

Để ΔABC~ΔA'B'C' thì \(\dfrac{AB}{A'B'}=\dfrac{AC}{A'C'}=\dfrac{BC}{B'C'}\)

Khi đó, hai tam giác đồng dạng với nhau theo trường hợp c-c-c

Câu 34:

ΔABC cân tại A

mà AD là đường phân giác

nên D là trung điểm của BC

=>\(BD=\dfrac{BC}{2}=\dfrac{10}{2}=5\left(cm\right)\)

=>Chọn A

Câu 35;

\(\dfrac{2x-1}{3}=\dfrac{x+2}{4}\)

=>4(2x-1)=3(x+2)

=>8x-4=3x+6

=>5x=10

=>x=2

=>Chọn D

Bạn tách nhỏ câu hỏi ra nhé

Bài 3:

a: \(\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)

=>\(\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}\right)-\left(x-23\right)\left(\dfrac{1}{26}+\dfrac{1}{27}\right)=0\)

=>\(\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)

=>x-23=0

=>x=23

b: \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

=>\(\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}=\dfrac{x+4+96}{96}+\dfrac{x+5+95}{95}\)

=>\(\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

=>x+100=0

=>x=-100

c: \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)

=>\(\left(\dfrac{x+1}{2004}+1\right)+\left(\dfrac{x+2}{2003}+1\right)=\left(\dfrac{x+3}{2002}+1\right)+\left(\dfrac{x+4}{2001}+1\right)\)

=>\(\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}=\dfrac{x+3+2002}{2002}+\dfrac{x+4+2001}{2001}\)

=>\(\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)

=>x+2005=0

=>x=-2005

d: Sửa đề:\(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}=-3\)

=>\(\left(\dfrac{201-x}{99}+1\right)+\left(\dfrac{203-x}{97}+1\right)+\left(\dfrac{205-x}{95}+1\right)=0\)

=>\(\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}+\dfrac{205-x+95}{95}=0\)

=>\(\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

=>300-x=0

=>x=300

Bài 1:

1: a: 15-8x=9-5x

=>-8x+5x=9-15

=>-3x=-6

=>x=2

b: 3+2x=5x+2

=>2x-5x=2-3

=>-3x=-1

=>\(x=\dfrac{1}{3}\)

c: \(5-\left(x-6\right)=4\left(3-2x\right)\)

=>\(5-x+6=12-8x\)

=>\(11-x=12-8x\)

=>\(-x+8x=12-11\)

=>7x=1

=>\(x=\dfrac{1}{7}\)

d: \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)

=>8x=-16

=>x=-2

2:

a: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)

=>\(\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(8x+6\right)}{36}\)

=>\(30x+9=36+32x+24\)

=>32x+60=30x+9

=>2x=-51

=>\(x=-\dfrac{51}{2}\)

b: \(\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)

=>\(\dfrac{5\left(7x-1\right)}{30}+\dfrac{60x}{30}=\dfrac{6\left(16-x\right)}{30}\)

=>5(7x-1)+60x=6(16-x)

=>35x-5+60x=96-6x

=>101x=101

=>x=1

c: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)

=>\(\dfrac{3\left(3x+2\right)-3x-1}{6}=\dfrac{10}{6}+\dfrac{12x}{6}\)

=>9x+6-3x-1=10+12x

=>12x+10=6x+5

=>6x=-5

=>\(x=-\dfrac{5}{6}\)

d: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)

=>\(\dfrac{6\left(x+4\right)}{30}+\dfrac{30\left(-x+4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

=>6x+24-30x+120=10x-15x+30

=>-24x+144=-5x+30

=>-19x=-114

=>x=6

Câu 32: C

Câu 33: B