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NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 9 2025
a: Ta có: \(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)
=>\(3x+x-\frac{9}{20}=-\frac{13}{40}\)
=>\(4x=-\frac{13}{40}+\frac{9}{20}=-\frac{13}{40}+\frac{18}{40}=\frac{5}{40}=\frac18\)
=>\(x=\frac18:4=\frac{1}{32}\)
b: \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)
=>\(x+\frac14x-2,5=-\frac{11}{20}\)
=>\(1,25x=-0,55+2,5=1,95\)
=>\(x=\frac{1.95}{1.25}=\frac{195}{125}=\frac{39}{25}\)
c: \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)
=>\(\frac35x+x+0,5=-\frac{13}{15}\)
=>\(\frac85x=-\frac{13}{15}-0,5=-\frac{26}{30}-\frac{15}{30}=-\frac{41}{30}\)
=>\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=\frac{-41}{6\cdot8}=-\frac{41}{48}\)
d: \(-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)
=>\(-\frac23x+4x-\frac67=\frac37\)
=>\(\frac{10}{3}x=\frac37+\frac67=\frac97\)
=>\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)
S
subjects
CTVHS
6 tháng 9 2025
bài 11: câu a:
\(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)
\(3x+x-\frac{9}{20}=-\frac{13}{40}\)
\(4x=-\frac{13}{40}+\frac{9}{20}\)
\(4x=-\frac{13}{40}+\frac{18}{40}\)
\(4x=\frac{5}{40}\)
\(4x=\frac18\)
\(x=\frac18:4=\frac18\cdot\frac14=\frac{1}{32}\)
b. \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)
\(x+\frac14x-2,5=-\frac{11}{20}\)
\(\frac54x-2,5=-\frac{11}{20}\)
\(\frac54x=-\frac{11}{20}+2,5\)
\(\frac54x=\frac{39}{20}\)
\(x=\frac{39}{20}:\frac54=\frac{39}{20}\cdot\frac45=\frac{39}{25}\)
c. \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)
\(\frac35x+x+0,5=-\frac{13}{15}\)
\(\frac85x+\frac12=-\frac{13}{15}\)
\(\frac85x=-\frac{13}{15}-\frac12\)
\(\frac85x=-\frac{41}{30}\)
\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=-\frac{41}{48}\)
\(d.-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)
\(-\frac23x+4x-\frac67=\frac{9}{21}\)
\(\frac{10}{3}x=\frac97\)
\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 9 2025
a: \(\left(-\frac54x+3,25\right)\left\lbrack\frac35-\left(-\frac52x\right)\right\rbrack=0\)
=>\(\left(\frac54x-\frac{13}{4}\right)\left(\frac52x+\frac35\right)=0\)
=>\(\left[\begin{array}{l}\frac54x-\frac{13}{4}=0\\ \frac52x+\frac35=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac54x=\frac{13}{4}\\ \frac52x=-\frac35\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{13}{4}:\frac54=\frac{13}{5}\\ x=-\frac35:\frac52=-\frac{6}{25}\end{array}\right.\)
b: \(\left(-\frac72x+1,75\right)\left\lbrack\frac45-\left(-\frac53x\right)\right\rbrack=0\)
=>\(\left[\begin{array}{l}-\frac72x+1,75=0\\ \frac45-\left(-\frac53x\right)=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}-\frac72x=-1,75=-\frac74\\ \frac53x=-\frac45\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{-7}{4}:\frac{-7}{2}=\frac24=\frac12\\ x=-\frac45:\frac53=-\frac45\cdot\frac35=-\frac{12}{25}\end{array}\right.\)
c: \(\left(x^2-4\right)\left(x+\frac27\right)=0\)
=>\(\left[\begin{array}{l}x^2-4=0\\ x+\frac27=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=4\\ x=-\frac27\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-2\\ x=-\frac27\end{array}\right.\)
d: \(\left(25-x^2\right)\left(5x-\frac59\right)=0\)
=>\(\left[\begin{array}{l}25-x^2=0\\ 5x-\frac59=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=25\\ 5x=\frac59\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-5\\ x=\frac19\end{array}\right.\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
4 tháng 10 2025
Bài 5:
a: \(\left|x\right|\ge0\forall x\)
=>\(7\left|x\right|\ge0\forall x\)
=>A=7|x|-98>=-98∀x
Dấu '=' xảy ra khi x=0
b: \(\left|5x-15\right|\ge0\forall x\)
=>\(-\frac34\left|5x-15\right|\le0\forall x\)
=>\(-\frac34\left|5x-15\right|+3\le3\forall x\)
Dấu '=' xảy ra khi 5x-15=0
=>5x=15
=>x=3
Bài 4:
a: \(\left|3x+1\right|-\frac12=0\)
=>\(\left|3x+1\right|=\frac12\)
=>\(\left[\begin{array}{l}3x+1=\frac12\\ 3x+1=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=\frac12-1=-\frac12\\ 3x=-\frac12-1=-\frac32\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac16\\ x=-\frac12\end{array}\right.\)
b: \(\left|2x-\frac25\right|=\left|5x-1\right|\)
=>\(\left[\begin{array}{l}5x-1=2x-\frac25\\ 5x-1=-2x+\frac25\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=-\frac25+1=\frac35\\ 7x=\frac25+1=\frac75\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac15\\ x=\frac15\end{array}\right.\)
=>\(x=\frac15\)
c: \(\left|2x-1\right|-4x=\frac12\)
=>\(\left|2x-1\right|=4x+\frac12\)
=>\(\begin{cases}4x+\frac12\ge0\\ \left(4x+\frac12\right)^2=\left(2x-1\right)^2\end{cases}\Rightarrow\begin{cases}4x\ge-\frac12\\ \left(4x+\frac12-2x+1\right)\left(4x+\frac12+2x-1\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-\frac18\\ \left(2x+\frac32\right)\left(6x-\frac12\right)=0\end{cases}\Rightarrow x=\frac{1}{12}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 9 2025
d: \(\left(-\frac34+\frac25\right):\frac37+\left(\frac35-\frac14\right):\frac37\)
\(=\left(-\frac34+\frac25+\frac35-\frac14\right):\frac37\)
\(=\left(1-1\right):\frac37=0\)
e: \(\frac59:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac59:\left(\frac{1}{15}-\frac23\right)\)
\(=\frac59:\left(\frac{2}{22}-\frac{5}{22}\right)+\frac59:\left(\frac{1}{15}-\frac{10}{15}\right)\)
\(=\frac59:\frac{-3}{22}+\frac59:\frac{-9}{15}\)
\(=\frac59\cdot\frac{-22}{3}+\frac59\cdot\frac{-5}{3}=\frac59\left(-\frac{22}{3}-\frac53\right)=\frac59\cdot\frac{-27}{3}=-5\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
25 tháng 9 2025
Ta có: \(\hat{BAa}=\hat{ABd}\left(=130^0\right)\)
mà hai góc này là hai góc ở vị trí so le trong
nên ab//cd
NL
Nguyễn Lê Phước Thịnh
CTVHS
25 tháng 9 2025
Ta có: \(\hat{eAb}=\hat{ABd}\left(=50^0\right)\)
mà hai góc này là hai góc ở vị trí đồng vị
nên ab//cd
NL
Nguyễn Lê Phước Thịnh
CTVHS
11 tháng 8 2025
Bài 2:
Qua B, kẻ tia BD nằm giữa hai tia BA và BC sao cho BD//Ax//Cz
ta có: BD//Ax
=>\(\hat{xAB}+\hat{ABD}=180^0\) (hai góc trong cùng phía)
=>\(\hat{ABD}=180^0-125^0=55^0\)
Ta có: BD//Cz
=>\(\hat{DBC}+\hat{BCz}=180^0\) (hai góc trong cùng phía)
=>\(\hat{DBC}=180^0-130^0=50^0\)
Ta có: tia BD nằm giữa hai tia BA và BC
=>\(\hat{ABC}=\hat{DBA}+\hat{DBC}\)
=>\(\hat{ABC}=55^0+50^0=105^0\)
Bài 3:
Ax//yy'
=>\(\hat{xAB}=\hat{yBA}\) (hai góc so le trong)
=>\(\hat{yBA}=50^0\)
Cz//yy'
=>\(\hat{yBC}=\hat{zCB}\) (hai góc so le trong)
=>\(\hat{yBC}=40^0\)
Ta có: tia By nằm giữa hai tia BA và BC
=>\(\hat{ABC}=\hat{yBA}+\hat{yBC}=40^0+50^0=90^0\)
Bài 4:
Qua B, kẻ tia BD nằm giữa hai tia BA và BC sao cho BD//Ax//Cz
BD//Ax
=>\(\hat{xAB}+\hat{ABD}=180^0\) (hai góc trong cùng phía)
=>\(\hat{ABD}=180^0-110^0=70^0\)
ta có; tia BD nằm giữa hai tia BA và BC
=>\(\hat{DBA}+\hat{DBC}=\hat{ABC}\)
=>\(\hat{DBC}=100^0-70^0=30^0\)
Ta có: \(\hat{DBC}=\hat{zCB}\left(=30^0\right)\)
mà hai góc này là hai góc ở vị trí so le trong
nên BD//Cz
Ta có: BD//Ax
BD//Cz
Do đó: Ax//Cz
NL
Nguyễn Lê Phước Thịnh
CTVHS
11 tháng 8 2025
a: a//b
=>\(\hat{A_1}=\hat{B_3}\) (hai góc so le trong)
mà \(\hat{A_1}=65^0\)
nên \(\hat{B_3}=65^0\)
b: Ta có: \(\hat{B}_3+\hat{B_2}=180^0\) (hai góc kề bù)
=>\(\hat{B_2}=180^0-65^0=115^0\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 8 2025
Giải:
a; \(\hat{A_1}\) = \(65^0\) (gt)
\(\hat{A_1}\) = \(\hat{A_3}\) = 65\(^0\)(đối đỉnh)
\(\hat{A_3}\) = \(\hat{B_3}\) = \(65^0\) (slt)
b; \(\hat{B_2}\) + \(\hat{B_3}\) = 180\(^0\) (hai góc kề bù)
\(\hat{B_2}\) = 180\(^0\) - \(\hat{B_3}\)
\(\hat{B_2}\) = 180\(^0\) - 65\(^0\) = 115\(^0\)
Vậy a; \(\hat{B}_3\) = 65\(^0\)
b; \(\hat{B_2}\) = 115\(^0\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
12 tháng 9 2025
Bài 3:
a: \(\frac{31}{15}>1;\frac{15}{31}<1\)
Do đó: \(\frac{31}{15}>\frac{15}{31}\)
=>\(\left(\frac{31}{15}\right)^{11}>\left(\frac{15}{31}\right)^{11}\)
b: \(\frac89<1\)
=>\(\left(\frac89\right)^{23}>\left(\frac89\right)^{25}\)
=>\(-\left(\frac89\right)^{23}<-\left(\frac89\right)^{25}\)
=>\(\left(-\frac89\right)^{23}<\left(-\frac89\right)^{25}\)
c: \(27^{40}=\left(27^2\right)^{20}=729^{20}\)
\(64^{60}=\left(64^3\right)^{20}=262144^{20}\)
mà 729<262144
nên \(27^{40}<64^{60}\)
Bài 2:
a: \(A=\frac{1}{10}-\frac{1}{10\cdot9}-\frac{1}{9\cdot8}-\cdots-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(=\frac{1}{10}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdots+\frac{1}{9\cdot10}\right)\)
\(=\frac{1}{10}-\left(1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}\right)\)
\(=\frac{1}{10}-\left(1-\frac{1}{10}\right)=\frac{1}{10}-\frac{9}{10}=-\frac{8}{10}=-\frac45\)
b: \(B=\frac13+\frac{1}{3^2}+\cdots+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)
=>\(3B=1+\frac13+\cdots+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
=>\(3B-B=1+\frac13+\cdots+\frac{1}{3^{98}}+\frac{1}{3^{99}}-\frac13-\frac{1}{3^2}-\cdots-\frac{1}{3^{100}}\)
=>\(2B=1-\frac{1}{3^{100}}=\frac{3^{100}-1}{3^{100}}\)
=>\(B=\frac{3^{100}-1}{2\cdot3^{100}}\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
29 tháng 9 2025
\(A=\frac89-\frac{1}{72}-\frac{1}{56}-\cdots-\frac16-\frac12\)
\(=\frac89-\left(\frac12+\frac16+\cdots+\frac{1}{72}\right)\)
\(=\frac89-\left(1-\frac12+\frac12-\frac13+\cdots+\frac18-\frac19\right)\)
\(=\frac89-\left(1-\frac19\right)=\frac89-\frac89=0\)
Bài 4A: Rút gọn biểu thức
a) (4x−1)(3x+2)−5x(x−3) -> =(12x2+8x−3x−2)−(5x2−15x) -> =12x2+5x−2−5x2+15x -> =(12x2−5x2)+(5x+15x)−2 -> =7x2+20x−2
b) (5x−2)(x+1)−2x(x2+x−3) -> =(5x2+5x−2x−2)−(2x3+2x2−6x) -> =5x2+3x−2−2x3−2x2+6x -> =−2x3+(5x2−2x2)+(3x+6x)−2 -> =−2x3+3x2+9x−2
c) (x+1)(2x−1)+x(x2−x+1) -> =(2x2−x+2x−1)+(x3−x2+x) -> =2x2+x−1+x3−x2+x -> =x3+(2x2−x2)+(x+x)−1 -> =x3+x2+2x−1
d) (3x2+x+2)⋅3−(2x+1)⋅2⋅(3+x) -> =(9x2+3x+6)−2⋅(2x+1)(x+3) -> =9x2+3x+6−2⋅(2x2+6x+x+3) -> =9x2+3x+6−2⋅(2x2+7x+3) -> =9x2+3x+6−4x2−14x−6 -> =(9x2−4x2)+(3x−14x)+(6−6) -> =5x2−11x
ờ hơi mờ nên như này nhé
4A
a) \(\left(\right. 4 x - 1 \left.\right) \left(\right. 3 x + 2 \left.\right) - 5 x \left(\right. x - 3 \left.\right)\)
b) \(\left(\right. 5 x - 2 \left.\right) \left(\right. x + 1 \left.\right) - 2 x \left(\right. - x^{2} + 3 \left.\right)\)
c) \(\left(\right. x + 1 \left.\right) \left(\right. 2 x - 1 \left.\right) + x \left(\right. x^{2} - 2 x + 1 \left.\right)\)
d) \(\left(\right. 3 x^{2} + x + 2 \left.\right) \cdot 3 - \left(\right. 2 x + 4 \left.\right) \cdot \left(\right. 3 + x \left.\right)\)
3B
a) \(3 x \left(\right. - x^{2} - 5 \left.\right) + 5 x \left(\right. x^{3} + 7 \left.\right) - 3 x^{2} \left(\right. x^{2} - x \left.\right) + 2 \left(\right. 4 - x \left.\right)\)
b) \(25 x - 4 \left(\right. 3 x - 1 \left.\right) + 7 x \left(\right. 5 - 2 x^{2} \left.\right)\)
c) \(4xc\left(\right.x^3-4x^2\left.\right)+2xc-2xc^5\)
\(= (6x^2 + 4x - 3x - 2) - (5x^2 - 15x)\)
\(= 6x^2 + x - 2 - 5x^2 + 15x\)
\(= x^2 + 16x - 2\)b) \((5x - 2)(x + 1) - 2x(x^2 + x - 3)\)
\(= (5x^2 + 5x - 2x - 2) - (2x^3 + 2x^2 - 6x)\)
\(= 5x^2 + 3x - 2 - 2x^3 - 2x^2 + 6x\)
\(= -2x^3 + 3x^2 + 9x - 2\)c) \((x + 1)(2x - 1) + x(x^2 - x + 1)\)
\(= (2x^2 - x + 2x - 1) + (x^3 - x^2 + x)\)
\(= 2x^2 + x - 1 + x^3 - x^2 + x\)
\(= x^3 + x^2 + 2x - 1\)d) \((3x^2 + x + 2) \cdot 3 - (2x + 1) \cdot 2(3 + x)\)
\(= 9x^2 + 3x + 6 - (2x + 1)(6 + 2x)\)
\(= 9x^2 + 3x + 6 - (12x + 4x^2 + 6 + 2x)\)
\(= 9x^2 + 3x + 6 - 4x^2 - 14x - 6\)
\(= 5x^2 - 11x\)Bài 3B: Rút gọn biểu thứca) \(3x(-x^2 - 5) + 5x(x^2 + 7) - 3x^2(x + 5)\)
\(= -3x^3 - 15x + 5x^3 + 35x - 3x^3 - 15x^2\)
\(= -x^3 - 15x^2 + 20x\)b) \(25x - 4(3x - 1) + 7x(5 - 2x^2)\)
\(= 25x - 12x + 4 + 35x - 14x^3\)
\(= -14x^3 + 48x + 4\)c) \(4x(x^3 - 4x^2) + 2x(2x^3 - 5x^2 + 7x + 1)\)
\(= 4x^4 - 16x^3 + 4x^4 - 10x^3 + 14x^2 + 2x\)
\(= 8x^4 - 26x^3 + 14x^2 + 2x\)
pls ✔
4A:
a: (4x-1)(3x+2)-5x(x-3)
\(=12x^2+8x-3x-2-5x^2+15x\)
\(=7x^2+20x-2\)
b: (5x-2)(x+1)-2x\(\left(x^2+x-3\right)\)
\(=5x^2+5x-2x-2-2x^3-2x^2+6x\)
\(=-2x^3+3x^2+9x-2\)
c: \(\left(x+1\right)\left(2x-1\right)+x\left(x^2-x+1\right)\)
\(=2x^2-x+2x-1+x^3-x^2+x\)
\(=x^3+x^2+2x-1\)
d: \(3\left(3x^2+x+2\right)-\left(2x+1\right)\cdot2\cdot\left(x+3\right)\)
\(=9x^2+3x+6-2\left(2x^2+6x+x+3\right)\)
\(=9x^2+3x+6-4x^2-14x-6=5x^2-11x\)
3B:
a: \(3x\left(-x^2-5\right)+5x\left(x^3+7\right)-3x^2\left(x^2-x+5\right)+2\left(4-x\right)\)
\(=-3x^3-15x+5x^4+35x-3x^4+3x^3-15x^2+8-2x\)
\(=2x^4-15x^2+18x+8\)
b: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)
\(=25x-12x+4+35x-14x^3=-14x^3+48x+4\)
c: \(4x\left(x^3-4x^2\right)+2x^2\left(2x^3-3x^2+7x+1\right)\)
\(=4x^4-16x^3+4x^5-6x^4+14x^3+2x^2\)
\(=4x^5-2x^4-2x^3+2x^2\)